Posted on by Kalkicode
Code Dynamic Programming

# Longest common subsequence of three sequences

In this article, we will discuss the problem of finding the longest common subsequence (LCS) of three sequences. We will explain the problem statement, provide a pseudocode algorithm with an explanation, and discuss the resultant output. The time complexity of the code will also be analyzed.

## Problem Statement

The problem is to find the longest common subsequence among three given sequences. A subsequence is a sequence that can be derived by deleting some or no elements from the original sequence without changing the order of the remaining elements. The task is to find the length of the longest common subsequence that is present in all three sequences.

## Pseudocode Algorithm

Let's walk through the pseudocode algorithm for finding the longest common subsequence of three sequences:

```function maxValue(a, b):
if a > b:
return a
else:
return b

function lengthOfLCP(text1, text2, text3):
m = length(text1)
n = length(text2)
o = length(text3)
dp[m][n][o]

for i from 0 to m:
for j from 0 to n:
for k from 0 to o:
if i = 0 or j = 0 or k = 0:
dp[i][j][k] = 1
else if text1[i-1] = text2[j-1] and text2[j-1] = text3[k-1]:
dp[i][j][k] = dp[i-1][j-1][k-1] + 1
else:
dp[i][j][k] = maxValue(dp[i-1][j][k], dp[i][j-1][k], dp[i][j][k-1])

print "Given text1: " + text1
print "Given text2: " + text2
print "Given text3: " + text3
print "Result: " + dp[m-1][n-1][o-1]

text1 = "pdfpofindther"
text2 = "pminoiwinters"
text3 = "codepokingtenter"
lengthOfLCP(text1, text2, text3)
```

The algorithm uses dynamic programming to solve the problem. It creates a three-dimensional array, `dp`, to store the lengths of the common subsequences. The algorithm iterates through the lengths of the three sequences and compares the characters at each position. If the characters are equal, it adds 1 to the length of the previous common subsequence. Otherwise, it takes the maximum length from the three adjacent positions in the `dp` array.

## Code Solution

``````/*
C program for
Longest common subsequence of three sequences
*/
#include <stdio.h>
#include <string.h>

int maxValue(int a, int b)
{
if (a > b)
{
return a;
}
return b;
}
void lengthOfLCP(char *text1, char *text2, char *text3)
{
// Get the length of text1 and text2
int m = strlen(text1);
int n = strlen(text2);
int o = strlen(text3);
int dp[m][n][o];
// Execute loop through by size of text1
for (int i = 0; i < m; ++i)
{
// Execute loop through by size of text2
for (int j = 0; j < n; ++j)
{
// Execute loop through by size of text3
for (int k = 0; k < o; ++k)
{
if (i == 0 || j == 0 || k == 0)
{
dp[i][j][k] = 1;
}
else if (text1[i - 1] == text2[j - 1] &&
text2[j - 1] == text3[k - 1])
{
// When i-1 and j-1 and k-1 is character are same
dp[i][j][k] = dp[i - 1][j - 1][k - 1] + 1;
}
else
{
// Get max value
dp[i][j][k] = maxValue(
maxValue(dp[i - 1][j][k], dp[i][j - 1][k]),
dp[i][j][k - 1]);
}
}
}
}
// Given string
printf("\n Given text1 : %s", text1);
printf("\n Given text2 : %s", text2);
printf("\n Given text3 : %s", text3);
// Display length of LCS
printf("\n Result : %d", dp[m - 1][n - 1][o - 1]);
}
int main(int argc, char
const *argv[])
{
char *text1 = "pdfpofindther";
char *text2 = "pminoiwinters";
char *text3 = "codepokingtenter";
// "pointer" Common sequences
// Result : 7
lengthOfLCP(text1, text2, text3);
return 0;
}``````

#### Output

`````` Given text1 : pdfpofindther
Given text2 : pminoiwinters
Given text3 : codepokingtenter
Result : 7``````
``````/*
Java program for
Longest common subsequence of three sequences
*/
public class LCP
{
public int maxValue(int a, int b)
{
if (a > b)
{
return a;
}
return b;
}
public void lengthOfLCP(String text1, String text2, String text3)
{
// Get the length of text1 and text2
int m = text1.length();
int n = text2.length();
int o = text3.length();
int[][][] dp = new int[m][n][o];
// Execute loop through by size of text1
for (int i = 0; i < m; ++i)
{
// Execute loop through by size of text2
for (int j = 0; j < n; ++j)
{
// Execute loop through by size of text3
for (int k = 0; k < o; ++k)
{
if (i == 0 || j == 0 || k == 0)
{
dp[i][j][k] = 1;
}
else if (text1.charAt(i - 1) == text2.charAt(j - 1) &&
text2.charAt(j - 1) == text3.charAt(k - 1))
{
// When i-1 and j-1 and k-1 is character are same
dp[i][j][k] = dp[i - 1][j - 1][k - 1] + 1;
}
else
{
// Get max value
dp[i][j][k] = maxValue(
maxValue(dp[i - 1][j][k], dp[i][j - 1][k]),
dp[i][j][k - 1]);
}
}
}
}
// Given string
System.out.print("\n Given text1 : " + text1);
System.out.print("\n Given text2 : " + text2);
System.out.print("\n Given text3 : " + text3);
// Display length of LCS
System.out.print("\n Result : " + dp[m - 1][n - 1][o - 1] );
}
public static void main(String[] args)
{
LCP task = new LCP();
String text1 = "pdfpofindther";
String text2 = "pminoiwinters";
String text3 = "codepokingtenter";
}
}``````

#### Output

`````` Given text1 : pdfpofindther
Given text2 : pminoiwinters
Given text3 : codepokingtenter
Result : 7``````
``````// Include header file
#include <iostream>
#include <string>

using namespace std;
/*
C++ program for
Longest common subsequence of three sequences
*/
class LCP
{
public: int maxValue(int a, int b)
{
if (a > b)
{
return a;
}
return b;
}
void lengthOfLCP(string text1, string text2, string text3)
{
// Get the length of text1 and text2
int m = text1.length();
int n = text2.length();
int o = text3.length();
int dp[m][n][o];
// Execute loop through by size of text1
for (int i = 0; i < m; ++i)
{
// Execute loop through by size of text2
for (int j = 0; j < n; ++j)
{
// Execute loop through by size of text3
for (int k = 0; k < o; ++k)
{
if (i == 0 || j == 0 || k == 0)
{
dp[i][j][k] = 1;
}
else if (text1[i - 1] == text2[j - 1] &&
text2[j - 1] == text3[k - 1])
{
// When i-1 and j-1 and k-1 is character are same
dp[i][j][k] = dp[i - 1][j - 1][k - 1] + 1;
}
else
{
// Get max value
dp[i][j][k] = this->maxValue(
this->maxValue(dp[i - 1][j][k], dp[i][j - 1][k]),
dp[i][j][k - 1]);
}
}
}
}
// Given string
cout << "\n Given text1 : " << text1;
cout << "\n Given text2 : " << text2;
cout << "\n Given text3 : " << text3;
// Display length of LCS
cout << "\n Result : " << dp[m - 1][n - 1][o - 1];
}
};
int main()
{
LCP *task = new LCP();
string text1 = "pdfpofindther";
string text2 = "pminoiwinters";
string text3 = "codepokingtenter";
return 0;
}``````

#### Output

`````` Given text1 : pdfpofindther
Given text2 : pminoiwinters
Given text3 : codepokingtenter
Result : 7``````
``````// Include namespace system
using System;
/*
Csharp program for
Longest common subsequence of three sequences
*/
public class LCP
{
public int maxValue(int a, int b)
{
if (a > b)
{
return a;
}
return b;
}
public void lengthOfLCP(String text1, String text2, String text3)
{
// Get the length of text1 and text2
int m = text1.Length;
int n = text2.Length;
int o = text3.Length;
int[,,] dp = new int[m,n,o];
// Execute loop through by size of text1
for (int i = 0; i < m; ++i)
{
// Execute loop through by size of text2
for (int j = 0; j < n; ++j)
{
// Execute loop through by size of text3
for (int k = 0; k < o; ++k)
{
if (i == 0 || j == 0 || k == 0)
{
dp[i,j,k] = 1;
}
else if (text1[i - 1] == text2[j - 1] &&
text2[j - 1] == text3[k - 1])
{
// When i-1 and j-1 and k-1 is character are same
dp[i,j,k] = dp[i - 1,j - 1,k - 1] + 1;
}
else
{
// Get max value
dp[i,j,k] = this.maxValue(
this.maxValue(dp[i - 1,j,k], dp[i,j - 1,k]),
dp[i,j,k - 1]);
}
}
}
}
// Given string
Console.Write("\n Given text1 : " + text1);
Console.Write("\n Given text2 : " + text2);
Console.Write("\n Given text3 : " + text3);
// Display length of LCS
Console.Write("\n Result : " + dp[m - 1,n - 1,o - 1]);
}
public static void Main(String[] args)
{
LCP task = new LCP();
String text1 = "pdfpofindther";
String text2 = "pminoiwinters";
String text3 = "codepokingtenter";
}
}``````

#### Output

`````` Given text1 : pdfpofindther
Given text2 : pminoiwinters
Given text3 : codepokingtenter
Result : 7``````
``````package main
import "fmt"
/*
Go program for
Longest common subsequence of three sequences
*/

func maxValue(a, b int) int {
if a > b {
return a
}
return b
}
func lengthOfLCP(text1, text2, text3 string) {
// Get the length of text1 and text2
var m int = len(text1)
var n int = len(text2)
var o int = len(text3)
var dp = make([][][]int,m);
for i := 0 ; i < m ; i++ {
dp[i] = make([][]int,n)
for j := 0 ; j < n ; j++ {
dp[i][j] = make([]int,o)
}
}

// Execute loop through by size of text1
for i := 0 ; i < m ; i++ {
// Execute loop through by size of text2
for j := 0 ; j < n ; j++ {
// Execute loop through by size of text3
for k := 0 ; k < o ; k++ {
if i == 0 || j == 0 || k == 0 {
dp[i][j][k] = 1
} else if text1[i - 1] == text2[j - 1] && text2[j - 1] == text3[k - 1] {
// When i-1 and j-1 and k-1 is character are same
dp[i][j][k] = dp[i - 1][j - 1][k - 1] + 1
} else {
// Get max value
dp[i][j][k] = maxValue(maxValue(dp[i - 1][j][k], dp[i][j - 1][k]), dp[i][j][k - 1])
}
}
}
}
// Given string
fmt.Print("\n Given text1 : ", text1)
fmt.Print("\n Given text2 : ", text2)
fmt.Print("\n Given text3 : ", text3)
// Display length of LCS
fmt.Print("\n Result : ", dp[m - 1][n - 1][o - 1])
}
func main() {

var text1 string = "pdfpofindther"
var text2 string = "pminoiwinters"
var text3 string = "codepokingtenter"
lengthOfLCP(text1, text2, text3)
}``````

#### Output

`````` Given text1 : pdfpofindther
Given text2 : pminoiwinters
Given text3 : codepokingtenter
Result : 7``````
``````<?php
/*
Php program for
Longest common subsequence of three sequences
*/
class LCP
{
public	function maxValue(\$a, \$b)
{
if (\$a > \$b)
{
return \$a;
}
return \$b;
}
public	function lengthOfLCP(\$text1, \$text2, \$text3)
{
// Get the length of text1 and text2
\$m = strlen(\$text1);
\$n = strlen(\$text2);
\$o = strlen(\$text3);
\$dp = array_fill(0, \$m, array_fill(0, \$n, array_fill(0, \$o, 0)));
// Execute loop through by size of text1
for (\$i = 0; \$i < \$m; ++\$i)
{
// Execute loop through by size of text2
for (\$j = 0; \$j < \$n; ++\$j)
{
// Execute loop through by size of text3
for (\$k = 0; \$k < \$o; ++\$k)
{
if (\$i == 0 || \$j == 0 || \$k == 0)
{
\$dp[\$i][\$j][\$k] = 1;
}
else if (\$text1[\$i - 1] == \$text2[\$j - 1] &&
\$text2[\$j - 1] == \$text3[\$k - 1])
{
// When i-1 and j-1 and k-1 is character are same
\$dp[\$i][\$j][\$k] = \$dp[\$i - 1][\$j - 1][\$k - 1] + 1;
}
else
{
// Get max value
\$dp[\$i][\$j][\$k] = \$this->maxValue(
\$this->maxValue(\$dp[\$i - 1][\$j][\$k],
\$dp[\$i][\$j - 1][\$k]),
\$dp[\$i][\$j][\$k - 1]);
}
}
}
}
// Given string
echo("\n Given text1 : ".\$text1);
echo("\n Given text2 : ".\$text2);
echo("\n Given text3 : ".\$text3);
// Display length of LCS
echo("\n Result : ".\$dp[\$m - 1][\$n - 1][\$o - 1]);
}
}

function main()
{
\$task = new LCP();
\$text1 = "pdfpofindther";
\$text2 = "pminoiwinters";
\$text3 = "codepokingtenter";
}
main();``````

#### Output

`````` Given text1 : pdfpofindther
Given text2 : pminoiwinters
Given text3 : codepokingtenter
Result : 7``````
``````/*
Node JS program for
Longest common subsequence of three sequences
*/
class LCP
{
maxValue(a, b)
{
if (a > b)
{
return a;
}
return b;
}
lengthOfLCP(text1, text2, text3)
{
// Get the length of text1 and text2
var m = text1.length;
var n = text2.length;
var o = text3.length;
var dp =
Array(m).fill(0).map(() => new Array(n).fill(0).map(
() => new Array(o).fill(0)));
// Execute loop through by size of text1
for (var i = 0; i < m; ++i)
{
// Execute loop through by size of text2
for (var j = 0; j < n; ++j)
{
// Execute loop through by size of text3
for (var k = 0; k < o; ++k)
{
if (i == 0 || j == 0 || k == 0)
{
dp[i][j][k] = 1;
}
else if (text1.charAt(i - 1) == text2.charAt(j - 1) &&
text2.charAt(j - 1) == text3.charAt(k - 1))
{
// When i-1 and j-1 and k-1 is character are same
dp[i][j][k] = dp[i - 1][j - 1][k - 1] + 1;
}
else
{
// Get max value
dp[i][j][k] = this.maxValue(
this.maxValue(dp[i - 1][j][k], dp[i][j - 1][k]),
dp[i][j][k - 1]);
}
}
}
}
// Given string
process.stdout.write("\n Given text1 : " + text1);
process.stdout.write("\n Given text2 : " + text2);
process.stdout.write("\n Given text3 : " + text3);
// Display length of LCS
process.stdout.write("\n Result : " + dp[m - 1][n - 1][o - 1]);
}
}

function main()
{
var task = new LCP();
var text1 = "pdfpofindther";
var text2 = "pminoiwinters";
var text3 = "codepokingtenter";
}
main();``````

#### Output

`````` Given text1 : pdfpofindther
Given text2 : pminoiwinters
Given text3 : codepokingtenter
Result : 7``````
``````#    Python 3 program for
#    Longest common subsequence of three sequences
class LCP :
def maxValue(self, a, b) :
if (a > b) :
return a

return b

def lengthOfLCP(self, text1, text2, text3) :
#  Get the length of text1 and text2
m = len(text1)
n = len(text2)
o = len(text3)
dp = [[[0] * (o) for _ in range(n) ]
for _ in range(m) ]
i = 0
#  Execute loop through by size of text1
while (i < m) :
j = 0
#  Execute loop through by size of text2
while (j < n) :
k = 0
#  Execute loop through by size of text3
while (k < o) :
if (i == 0 or j == 0 or k == 0) :
dp[i][j][k] = 1
elif (text1[i - 1] == text2[j - 1] and
text2[j - 1] == text3[k - 1]) :
#  When i-1 and j-1 and k-1 is character are same
dp[i][j][k] = dp[i - 1][j - 1][k - 1] + 1
else :
#  Get max value
dp[i][j][k] = self.maxValue(
self.maxValue(dp[i - 1][j][k], dp[i][j - 1][k]),
dp[i][j][k - 1])

k += 1

j += 1

i += 1

#  Given string
print("\n Given text1 : ", text1, end = "")
print("\n Given text2 : ", text2, end = "")
print("\n Given text3 : ", text3, end = "")
#  Display length of LCS
print("\n Result : ", dp[m - 1][n - 1][o - 1], end = "")

def main() :
text1 = "pdfpofindther"
text2 = "pminoiwinters"
text3 = "codepokingtenter"

if __name__ == "__main__": main()``````

#### Output

`````` Given text1 :  pdfpofindther
Given text2 :  pminoiwinters
Given text3 :  codepokingtenter
Result :  7``````
``````#    Ruby program for
#    Longest common subsequence of three sequences
class LCP
def maxValue(a, b)
if (a > b)
return a
end

return b
end

def lengthOfLCP(text1, text2, text3)
#  Get the length of text1 and text2
m = text1.length
n = text2.length
o = text3.length
dp = Array.new(m) {Array.new(n) {Array.new(o) {0}}}
i = 0
#  Execute loop through by size of text1
while (i < m)
j = 0
#  Execute loop through by size of text2
while (j < n)
k = 0
#  Execute loop through by size of text3
while (k < o)
if (i == 0 || j == 0 || k == 0)
dp[i][j][k] = 1
elsif (text1[i - 1] == text2[j - 1] &&
text2[j - 1] == text3[k - 1])
#  When i-1 and j-1 and k-1 is character are same
dp[i][j][k] = dp[i - 1][j - 1][k - 1] + 1
else

#  Get max value
dp[i][j][k] = self.maxValue(
self.maxValue(dp[i - 1][j][k], dp[i][j - 1][k]),
dp[i][j][k - 1])
end

k += 1
end

j += 1
end

i += 1
end

#  Given string
print("\n Given text1 : ", text1)
print("\n Given text2 : ", text2)
print("\n Given text3 : ", text3)
#  Display length of LCS
print("\n Result : ", dp[m - 1][n - 1][o - 1])
end

end

def main()
text1 = "pdfpofindther"
text2 = "pminoiwinters"
text3 = "codepokingtenter"
end

main()``````

#### Output

`````` Given text1 : pdfpofindther
Given text2 : pminoiwinters
Given text3 : codepokingtenter
Result : 7``````
``````/*
Scala program for
Longest common subsequence of three sequences
*/
class LCP()
{
def maxValue(a: Int, b: Int): Int = {
if (a > b)
{
return a;
}
return b;
}
def lengthOfLCP(text1: String, text2: String, text3: String): Unit = {
// Get the length of text1 and text2
var m: Int = text1.length();
var n: Int = text2.length();
var o: Int = text3.length();
var dp: Array[Array[Array[Int]]] = Array.fill[Int](m, n, o)(0);
var i: Int = 0;
// Execute loop through by size of text1
while (i < m)
{
var j: Int = 0;
// Execute loop through by size of text2
while (j < n)
{
var k: Int = 0;
// Execute loop through by size of text3
while (k < o)
{
if (i == 0 || j == 0 || k == 0)
{
dp(i)(j)(k) = 1;
}
else if (text1.charAt(i - 1) == text2.charAt(j - 1) &&
text2.charAt(j - 1) == text3.charAt(k - 1))
{
// When i-1 and j-1 and k-1 is character are same
dp(i)(j)(k) = dp(i - 1)(j - 1)(k - 1) + 1;
}
else
{
// Get max value
dp(i)(j)(k) = maxValue(
maxValue(dp(i - 1)(j)(k), dp(i)(j - 1)(k)),
dp(i)(j)(k - 1));
}
k += 1;
}
j += 1;
}
i += 1;
}
// Given string
print("\n Given text1 : " + text1);
print("\n Given text2 : " + text2);
print("\n Given text3 : " + text3);
// Display length of LCS
print("\n Result : " + dp(m - 1)(n - 1)(o - 1));
}
}
object Main
{
def main(args: Array[String]): Unit = {
var task: LCP = new LCP();
var text1: String = "pdfpofindther";
var text2: String = "pminoiwinters";
var text3: String = "codepokingtenter";
}
}``````

#### Output

`````` Given text1 : pdfpofindther
Given text2 : pminoiwinters
Given text3 : codepokingtenter
Result : 7``````
``````import Foundation;
/*
Swift 4 program for
Longest common subsequence of three sequences
*/
class LCP
{
func maxValue(_ a: Int, _ b: Int) -> Int
{
if (a > b)
{
return a;
}
return b;
}
func lengthOfLCP(_ a: String, _ b: String, _ c: String)
{
// Get the length of text1 and text2
let m: Int = a.count;
let n: Int = b.count;
let o: Int = c.count;
let text1 = Array(a);
let text2 = Array(b);
let text3 = Array(c);

var dp: [[[Int]]] = Array(
repeating: Array(
repeating: Array(repeating: 0, count: o), count: n), count: m);
var i: Int = 0;
// Execute loop through by size of text1
while (i < m)
{
var j: Int = 0;
// Execute loop through by size of text2
while (j < n)
{
var k: Int = 0;
// Execute loop through by size of text3
while (k < o)
{
if (i == 0 || j == 0 || k == 0)
{
dp[i][j][k] = 1;
}
else if (text1[i - 1] == text2[j - 1] &&
text2[j - 1] == text3[k - 1])
{
// When i-1 and j-1 and k-1 is character are same
dp[i][j][k] = dp[i - 1][j - 1][k - 1] + 1;
}
else
{
// Get max value
dp[i][j][k] = self.maxValue(
self.maxValue(dp[i - 1][j][k], dp[i][j - 1][k]),
dp[i][j][k - 1]);
}
k += 1;
}
j += 1;
}
i += 1;
}
// Given string
print("\n Given text1 : ", a, terminator: "");
print("\n Given text2 : ", b, terminator: "");
print("\n Given text3 : ", c, terminator: "");
// Display length of LCS
print("\n Result : ", dp[m - 1][n - 1][o - 1], terminator: "");
}
}
func main()
{
let task: LCP = LCP();
let text1: String = "pdfpofindther";
let text2: String = "pminoiwinters";
let text3: String = "codepokingtenter";
}
main();``````

#### Output

`````` Given text1 :  pdfpofindther
Given text2 :  pminoiwinters
Given text3 :  codepokingtenter
Result :  7``````
``````/*
Kotlin program for
Longest common subsequence of three sequences
*/
class LCP
{
fun maxValue(a: Int, b: Int): Int
{
if (a > b)
{
return a;
}
return b;
}
fun lengthOfLCP(text1: String, text2: String, text3: String): Unit
{
// Get the length of text1 and text2
val m: Int = text1.length;
val n: Int = text2.length;
val o: Int = text3.length;
var dp: Array < Array < Array < Int >>> = Array(m)
{
Array(n)
{
Array(o)
{
0
}
}
};
var i: Int = 0;
// Execute loop through by size of text1
while (i < m)
{
var j: Int = 0;
// Execute loop through by size of text2
while (j < n)
{
var k: Int = 0;
// Execute loop through by size of text3
while (k < o)
{
if (i == 0 || j == 0 || k == 0)
{
dp[i][j][k] = 1;
}
else if (text1.get(i - 1) == text2.get(j - 1) &&
text2.get(j - 1) == text3.get(k - 1))
{
// When i-1 and j-1 and k-1 is character are same
dp[i][j][k] = dp[i - 1][j - 1][k - 1] + 1;
}
else
{
// Get max value
dp[i][j][k] = this.maxValue(
this.maxValue(dp[i - 1][j][k], dp[i][j - 1][k]),
dp[i][j][k - 1]);
}
k += 1;
}
j += 1;
}
i += 1;
}
// Given string
print("\n Given text1 : " + text1);
print("\n Given text2 : " + text2);
print("\n Given text3 : " + text3);
// Display length of LCS
print("\n Result : " + dp[m - 1][n - 1][o - 1]);
}
}
fun main(args: Array < String > ): Unit
{
val task: LCP = LCP();
val text1: String = "pdfpofindther";
val text2: String = "pminoiwinters";
val text3: String = "codepokingtenter";
}``````

#### Output

`````` Given text1 : pdfpofindther
Given text2 : pminoiwinters
Given text3 : codepokingtenter
Result : 7``````

## Resultant Output

When the given sequences are "pdfpofindther", "pminoiwinters", and "codepokingtenter", the output will be:

```Given text1: pdfpofindther
Given text2: pminoiwinters
Given text3: codepokingtenter
Result: 7
```

The output indicates that the longest common subsequence among the three sequences has a length of 7.

## Time Complexity

The time complexity of the algorithm is determined by the nested loops that iterate through the lengths of the three sequences. Let `m`, `n`, and `o` be the lengths of the three sequences, respectively.

The algorithm has a time complexity of `O(m * n * o)` because each element in the three-dimensional `dp` array is calculated once in constant time. Therefore, the algorithm's time complexity is directly proportional to the product of the lengths of the three sequences.

As a result, the algorithm efficiently finds the longest common subsequence among three sequences using dynamic programming.

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