Find the repeating and the missing element using xor

Here given code implementation process.

// C program
// Find the repeating and the missing element using xor
#include <stdio.h>

// Print the given array elements
void printArray(int arr[], int n)
{
	printf("\n Array Element");
	for (int i = 0; i < n; ++i)
	{
		printf("  %d", arr[i]);
	}
}
void repeatingAndMissing(int arr[], int n)
{
	if (n < 2)
	{
		return;
	}
	// Define auxiliary variables
	int repeated = 0;
	int missing = 0;
	int i = 0;
	int x_or = arr[i];
	int rightMost = 0;
	// Perform xor operation in given array elements
	for (i = 1; i < n; ++i)
	{
		x_or = x_or ^ arr[i];
	}
	// Perform xor operation in range from 1 to n
	for (i = 1; i <= n; ++i)
	{
		x_or = x_or ^ 1;
	}
	// Find rightmost set bits
	rightMost = ~(x_or - 1) & x_or;
	for (i = 0; i < n; ++i)
	{
		if ((arr[i] & rightMost) == 0)
		{
			// When element is repeated in array
			repeated = repeated ^ arr[i];
		}
		else
		{
			// When element is missing in array
			missing = missing ^ arr[i];
		}
	}
	for (i = 1; i <= n; ++i)
	{
		if ((i & rightMost) == 0)
		{
			// When element is repeated in range (1...n)
			repeated = repeated ^ i;
		}
		else
		{
			// When element is missing in range (1...n)
			missing = missing ^ i;
		}
	}
	printArray(arr, n);
	// Display calculated result
	printf("\n Missing  : %d", missing);
	printf("\n Repeated : %d\n", repeated);
}
int main()
{
	int arr1[] = {
		1 , 8 , 5 , 7 , 8 , 2 , 6 , 3
	};
	// Get the size of array arr1
	int n = sizeof(arr1) / sizeof(arr1[0]);
	repeatingAndMissing(arr1, n);
	int arr2[] = {
		1 , 4 , 3 , 3
	};
	// Get the size of array arr2
	n = sizeof(arr2) / sizeof(arr2[0]);
	repeatingAndMissing(arr2, n);
	return 0;
}

input

 Array Element  1  8  5  7  8  2  6  3
 Missing  : 4
 Repeated : 8

 Array Element  1  4  3  3
 Missing  : 3
 Repeated : 2
/*
  Java Program for
  Find the repeating and the missing element using xor
*/
class Finding
{
	// Print the given array elements
	public void printArray(int[] arr, int n)
	{
		System.out.print("\n Array Element");
		for (int i = 0; i < n; ++i)
		{
			System.out.print(" " + arr[i]);
		}
	}
	public void repeatingAndMissing(int[] arr, int n)
	{
		if (n < 2)
		{
			return;
		}
		// Define auxiliary variables
		int repeated = 0;
		int missing = 0;
		int i = 0;
		int x_or = arr[i];
		int rightMost = 0;
		// Perform xor operation in given array elements
		for (i = 1; i < n; ++i)
		{
			x_or = x_or ^ arr[i];
		}
		// Perform xor operation in range from 1 to n
		for (i = 1; i <= n; ++i)
		{
			x_or = x_or ^ 1;
		}
		// Find rightmost set bits
		rightMost = ~(x_or - 1) & x_or;
		for (i = 0; i < n; ++i)
		{
			if ((arr[i] & rightMost) == 0)
			{
				// When element is repeated in array
				repeated = repeated ^ arr[i];
			}
			else
			{
				// When element is missing in array
				missing = missing ^ arr[i];
			}
		}
		for (i = 1; i <= n; ++i)
		{
			if ((i & rightMost) == 0)
			{
				// When element is repeated in range (1...n)
				repeated = repeated ^ i;
			}
			else
			{
				// When element is missing in range (1...n)
				missing = missing ^ i;
			}
		}
		printArray(arr, n);
		// Display calculated result
		System.out.println("\n Missing : " + missing);
		System.out.println(" Repeated : " + repeated);
	}
	public static void main(String[] args)
	{
		Finding task = new Finding();
		int[] arr1 = {
			1 , 8 , 5 , 7 , 8 , 2 , 6 , 3
		};
		// Get the size of array arr1
		int n = arr1.length;
		task.repeatingAndMissing(arr1, n);
		int[] arr2 = {
			1 , 4 , 3 , 3
		};
		// Get the size of array arr2
		n = arr2.length;
		task.repeatingAndMissing(arr2, n);
	}
}

input

 Array Element 1 8 5 7 8 2 6 3
 Missing : 4
 Repeated : 8

 Array Element 1 4 3 3
 Missing : 3
 Repeated : 2
// Include header file
#include <iostream>

using namespace std;
/*
  C++ Program for
  Find the repeating and the missing element using xor
*/
class Finding
{
	public:
		// Print the given array elements
		void printArray(int arr[], int n)
		{
			cout << "\n Array Element";
			for (int i = 0; i < n; ++i)
			{
				cout << " " << arr[i];
			}
		}
	void repeatingAndMissing(int arr[], int n)
	{
		if (n < 2)
		{
			return;
		}
		// Define auxiliary variables
		int repeated = 0;
		int missing = 0;
		int i = 0;
		int x_or = arr[i];
		int rightMost = 0;
		// Perform xor operation in given array elements
		for (i = 1; i < n; ++i)
		{
			x_or = x_or ^ arr[i];
		}
		// Perform xor operation in range from 1 to n
		for (i = 1; i <= n; ++i)
		{
			x_or = x_or ^ 1;
		}
		// Find rightmost set bits
		rightMost = ~(x_or - 1) &x_or;
		for (i = 0; i < n; ++i)
		{
			if ((arr[i] &rightMost) == 0)
			{
				// When element is repeated in array
				repeated = repeated ^ arr[i];
			}
			else
			{
				// When element is missing in array
				missing = missing ^ arr[i];
			}
		}
		for (i = 1; i <= n; ++i)
		{
			if ((i &rightMost) == 0)
			{
				// When element is repeated in range (1...n)
				repeated = repeated ^ i;
			}
			else
			{
				// When element is missing in range (1...n)
				missing = missing ^ i;
			}
		}
		this->printArray(arr, n);
		// Display calculated result
		cout << "\n Missing : " << missing << endl;
		cout << " Repeated : " << repeated << endl;
	}
};
int main()
{
	Finding *task = new Finding();
	int arr1[] = {
		1 , 8 , 5 , 7 , 8 , 2 , 6 , 3
	};
	// Get the size of array arr1
	int n = sizeof(arr1) / sizeof(arr1[0]);
	task->repeatingAndMissing(arr1, n);
	int arr2[] = {
		1 , 4 , 3 , 3
	};
	// Get the size of array arr2
	n = sizeof(arr2) / sizeof(arr2[0]);
	task->repeatingAndMissing(arr2, n);
	return 0;
}

input

 Array Element 1 8 5 7 8 2 6 3
 Missing : 4
 Repeated : 8

 Array Element 1 4 3 3
 Missing : 3
 Repeated : 2
// Include namespace system
using System;
/*
  Csharp Program for
  Find the repeating and the missing element using xor
*/
public class Finding
{
	// Print the given array elements
	public void printArray(int[] arr, int n)
	{
		Console.Write("\n Array Element");
		for (int i = 0; i < n; ++i)
		{
			Console.Write(" " + arr[i]);
		}
	}
	public void repeatingAndMissing(int[] arr, int n)
	{
		if (n < 2)
		{
			return;
		}
		// Define auxiliary variables
		int repeated = 0;
		int missing = 0;
		int i = 0;
		int x_or = arr[i];
		int rightMost = 0;
		// Perform xor operation in given array elements
		for (i = 1; i < n; ++i)
		{
			x_or = x_or ^ arr[i];
		}
		// Perform xor operation in range from 1 to n
		for (i = 1; i <= n; ++i)
		{
			x_or = x_or ^ 1;
		}
		// Find rightmost set bits
		rightMost = ~(x_or - 1) & x_or;
		for (i = 0; i < n; ++i)
		{
			if ((arr[i] & rightMost) == 0)
			{
				// When element is repeated in array
				repeated = repeated ^ arr[i];
			}
			else
			{
				// When element is missing in array
				missing = missing ^ arr[i];
			}
		}
		for (i = 1; i <= n; ++i)
		{
			if ((i & rightMost) == 0)
			{
				// When element is repeated in range (1...n)
				repeated = repeated ^ i;
			}
			else
			{
				// When element is missing in range (1...n)
				missing = missing ^ i;
			}
		}
		this.printArray(arr, n);
		// Display calculated result
		Console.WriteLine("\n Missing : " + missing);
		Console.WriteLine(" Repeated : " + repeated);
	}
	public static void Main(String[] args)
	{
		Finding task = new Finding();
		int[] arr1 = {
			1 , 8 , 5 , 7 , 8 , 2 , 6 , 3
		};
		// Get the size of array arr1
		int n = arr1.Length;
		task.repeatingAndMissing(arr1, n);
		int[] arr2 = {
			1 , 4 , 3 , 3
		};
		// Get the size of array arr2
		n = arr2.Length;
		task.repeatingAndMissing(arr2, n);
	}
}

input

 Array Element 1 8 5 7 8 2 6 3
 Missing : 4
 Repeated : 8

 Array Element 1 4 3 3
 Missing : 3
 Repeated : 2
<?php
/*
  Php Program for
  Find the repeating and the missing element using xor
*/
class Finding
{
	// Print the given array elements
	public	function printArray($arr, $n)
	{
		echo "\n Array Element";
		for ($i = 0; $i < $n; ++$i)
		{
			echo " ".$arr[$i];
		}
	}
	public	function repeatingAndMissing($arr, $n)
	{
		if ($n < 2)
		{
			return;
		}
		// Define auxiliary variables
		$repeated = 0;
		$missing = 0;
		$i = 0;
		$x_or = $arr[$i];
		$rightMost = 0;
		// Perform xor operation in given array elements
		for ($i = 1; $i < $n; ++$i)
		{
			$x_or = $x_or ^ $arr[$i];
		}
		// Perform xor operation in range from 1 to n
		for ($i = 1; $i <= $n; ++$i)
		{
			$x_or = $x_or ^ 1;
		}
		// Find rightmost set bits
		$rightMost = ~($x_or - 1) & $x_or;
		for ($i = 0; $i < $n; ++$i)
		{
			if (($arr[$i] & $rightMost) == 0)
			{
				// When element is repeated in array
				$repeated = $repeated ^ $arr[$i];
			}
			else
			{
				// When element is missing in array
				$missing = $missing ^ $arr[$i];
			}
		}
		for ($i = 1; $i <= $n; ++$i)
		{
			if (($i & $rightMost) == 0)
			{
				// When element is repeated in range (1...n)
				$repeated = $repeated ^ $i;
			}
			else
			{
				// When element is missing in range (1...n)
				$missing = $missing ^ $i;
			}
		}
		$this->printArray($arr, $n);
		// Display calculated result
		echo "\n Missing : ".$missing.
		"\n";
		echo " Repeated : ".$repeated.
		"\n";
	}
}

function main()
{
	$task = new Finding();
	$arr1 = array(1, 8, 5, 7, 8, 2, 6, 3);
	// Get the size of array arr1
	$n = count($arr1);
	$task->repeatingAndMissing($arr1, $n);
	$arr2 = array(1, 4, 3, 3);
	// Get the size of array arr2
	$n = count($arr2);
	$task->repeatingAndMissing($arr2, $n);
}
main();

input

 Array Element 1 8 5 7 8 2 6 3
 Missing : 4
 Repeated : 8

 Array Element 1 4 3 3
 Missing : 3
 Repeated : 2
/*
  Node JS Program for
  Find the repeating and the missing element using xor
*/
class Finding
{
	// Print the given array elements
	printArray(arr, n)
	{
		process.stdout.write("\n Array Element");
		for (var i = 0; i < n; ++i)
		{
			process.stdout.write(" " + arr[i]);
		}
	}
	repeatingAndMissing(arr, n)
	{
		if (n < 2)
		{
			return;
		}
		// Define auxiliary variables
		var repeated = 0;
		var missing = 0;
		var i = 0;
		var x_or = arr[i];
		var rightMost = 0;
		// Perform xor operation in given array elements
		for (i = 1; i < n; ++i)
		{
			x_or = x_or ^ arr[i];
		}
		// Perform xor operation in range from 1 to n
		for (i = 1; i <= n; ++i)
		{
			x_or = x_or ^ 1;
		}
		// Find rightmost set bits
		rightMost = ~(x_or - 1) & x_or;
		for (i = 0; i < n; ++i)
		{
			if ((arr[i] & rightMost) == 0)
			{
				// When element is repeated in array
				repeated = repeated ^ arr[i];
			}
			else
			{
				// When element is missing in array
				missing = missing ^ arr[i];
			}
		}
		for (i = 1; i <= n; ++i)
		{
			if ((i & rightMost) == 0)
			{
				// When element is repeated in range (1...n)
				repeated = repeated ^ i;
			}
			else
			{
				// When element is missing in range (1...n)
				missing = missing ^ i;
			}
		}
		this.printArray(arr, n);
		// Display calculated result
		console.log("\n Missing : " + missing);
		console.log(" Repeated : " + repeated);
	}
}

function main()
{
	var task = new Finding();
	var arr1 = [1, 8, 5, 7, 8, 2, 6, 3];
	// Get the size of array arr1
	var n = arr1.length;
	task.repeatingAndMissing(arr1, n);
	var arr2 = [1, 4, 3, 3];
	// Get the size of array arr2
	n = arr2.length;
	task.repeatingAndMissing(arr2, n);
}
main();

input

 Array Element 1 8 5 7 8 2 6 3
 Missing : 4
 Repeated : 8

 Array Element 1 4 3 3
 Missing : 3
 Repeated : 2
#  Python 3 Program for
#  Find the repeating and the missing element using xor
class Finding :
	#  Print the given list elements
	def printArray(self, arr, n) :
		print("\n Array Element", end = "")
		i = 0
		while (i < n) :
			print(" ", arr[i], end = "")
			i += 1
		
	
	def repeatingAndMissing(self, arr, n) :
		if (n < 2) :
			return
		
		repeated = 0
		missing = 0
		i = 0
		x_or = arr[i]
		rightMost = 0
		#  Perform xor operation in given list elements
		i = 1
		while (i < n) :
			x_or = x_or ^ arr[i]
			i += 1
		
		#  Perform xor operation in range from 1 to n
		i = 1
		while (i <= n) :
			x_or = x_or ^ 1
			i += 1
		
		#  Find rightmost set bits
		rightMost = ~(x_or - 1) & x_or
		i = 0
		while (i < n) :
			if ((arr[i] & rightMost) == 0) :
				#  When element is repeated in list
				repeated = repeated ^ arr[i]
			else :
				#  When element is missing in list
				missing = missing ^ arr[i]
			
			i += 1
		
		i = 1
		while (i <= n) :
			if ((i & rightMost) == 0) :
				#  When element is repeated in range (1...n)
				repeated = repeated ^ i
			else :
				#  When element is missing in range (1...n)
				missing = missing ^ i
			
			i += 1
		
		self.printArray(arr, n)
		#  Display calculated result
		print("\n Missing : ", missing)
		print(" Repeated : ", repeated)
	

def main() :
	task = Finding()
	arr1 = [1, 8, 5, 7, 8, 2, 6, 3]
	n = len(arr1)
	task.repeatingAndMissing(arr1, n)
	arr2 = [1, 4, 3, 3]
	#  Get the size of list arr2
	n = len(arr2)
	task.repeatingAndMissing(arr2, n)

if __name__ == "__main__": main()

input

 Array Element  1  8  5  7  8  2  6  3
 Missing :  4
 Repeated :  8

 Array Element  1  4  3  3
 Missing :  3
 Repeated :  2
#  Ruby Program for
#  Find the repeating and the missing element using xor
class Finding 
	#  Print the given array elements
	def printArray(arr, n) 
		print("\n Array Element")
		i = 0
		while (i < n) 
			print(" ", arr[i])
			i += 1
		end

	end

	def repeatingAndMissing(arr, n) 
		if (n < 2) 
			return
		end

		#  Define auxiliary variables
		repeated = 0
		missing = 0
		i = 0
		x_or = arr[i]
		rightMost = 0
		#  Perform xor operation in given array elements
		i = 1
		while (i < n) 
			x_or = x_or ^ arr[i]
			i += 1
		end

		#  Perform xor operation in range from 1 to n
		i = 1
		while (i <= n) 
			x_or = x_or ^ 1
			i += 1
		end

		#  Find rightmost set bits
		rightMost = ~(x_or - 1) & x_or
		i = 0
		while (i < n) 
			if ((arr[i] & rightMost) == 0) 
				#  When element is repeated in array
				repeated = repeated ^ arr[i]
			else 
				#  When element is missing in array
				missing = missing ^ arr[i]
			end

			i += 1
		end

		i = 1
		while (i <= n) 
			if ((i & rightMost) == 0) 
				#  When element is repeated in range (1...n)
				repeated = repeated ^ i
			else 
				#  When element is missing in range (1...n)
				missing = missing ^ i
			end

			i += 1
		end

		self.printArray(arr, n)
		#  Display calculated result
		print("\n Missing : ", missing, "\n")
		print(" Repeated : ", repeated, "\n")
	end

end

def main() 
	task = Finding.new()
	arr1 = [1, 8, 5, 7, 8, 2, 6, 3]
	#  Get the size of array arr1
	n = arr1.length
	task.repeatingAndMissing(arr1, n)
	arr2 = [1, 4, 3, 3]
	#  Get the size of array arr2
	n = arr2.length
	task.repeatingAndMissing(arr2, n)
end

main()

input

 Array Element 1 8 5 7 8 2 6 3
 Missing : 4
 Repeated : 8

 Array Element 1 4 3 3
 Missing : 3
 Repeated : 2
/*
  Scala Program for
  Find the repeating and the missing element using xor
*/
class Finding()
{
	// Print the given array elements
	def printArray(arr: Array[Int], n: Int): Unit = {
		print("\n Array Element");
		var i: Int = 0;
		while (i < n)
		{
			print(" " + arr(i));
			i += 1;
		}
	}
	def repeatingAndMissing(arr: Array[Int], n: Int): Unit = {
		if (n < 2)
		{
			return;
		}
		// Define auxiliary variables
		var repeated: Int = 0;
		var missing: Int = 0;
		var i: Int = 0;
		var x_or: Int = arr(i);
		var rightMost: Int = 0;
		// Perform xor operation in given array elements
		i = 1;
		while (i < n)
		{
			x_or = x_or ^ arr(i);
			i += 1;
		}
		// Perform xor operation in range from 1 to n
		i = 1;
		while (i <= n)
		{
			x_or = x_or ^ 1;
			i += 1;
		}
		// Find rightmost set bits
		rightMost = ~(x_or - 1) & x_or;
		i = 0;
		while (i < n)
		{
			if ((arr(i) & rightMost) == 0)
			{
				// When element is repeated in array
				repeated = repeated ^ arr(i);
			}
			else
			{
				// When element is missing in array
				missing = missing ^ arr(i);
			}
			i += 1;
		}
		i = 1;
		while (i <= n)
		{
			if ((i & rightMost) == 0)
			{
				// When element is repeated in range (1...n)
				repeated = repeated ^ i;
			}
			else
			{
				// When element is missing in range (1...n)
				missing = missing ^ i;
			}
			i += 1;
		}
		printArray(arr, n);
		// Display calculated result
		println("\n Missing : " + missing);
		println(" Repeated : " + repeated);
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Finding = new Finding();
		var arr1: Array[Int] = Array(1, 8, 5, 7, 8, 2, 6, 3);
		// Get the size of array arr1
		var n: Int = arr1.length;
		task.repeatingAndMissing(arr1, n);
		var arr2: Array[Int] = Array(1, 4, 3, 3);
		// Get the size of array arr2
		n = arr2.length;
		task.repeatingAndMissing(arr2, n);
	}
}

input

 Array Element 1 8 5 7 8 2 6 3
 Missing : 4
 Repeated : 8

 Array Element 1 4 3 3
 Missing : 3
 Repeated : 2
/*
  Swift 4 Program for
  Find the repeating and the missing element using xor
*/
class Finding
{
	// Print the given array elements
	func printArray(_ arr: [Int], _ n: Int)
	{
		print("\n Array Element", terminator: "");
		var i: Int = 0;
		while (i < n)
		{
			print(" ", arr[i], terminator: "");
			i += 1;
		}
	}
	func repeatingAndMissing(_ arr: [Int], _ n: Int)
	{
		if (n < 2)
		{
			return;
		}
		// Define auxiliary variables
		var repeated: Int = 0;
		var missing: Int = 0;
		var i: Int = 0;
		var x_or: Int = arr[i];
		var rightMost: Int = 0;
		// Perform xor operation in given array elements
		i = 1;
		while (i < n)
		{
			x_or = x_or ^ arr[i];
			i += 1;
		}
		// Perform xor operation in range from 1 to n
		i = 1;
		while (i <= n)
		{
			x_or = x_or ^ 1;
			i += 1;
		}
		// Find rightmost set bits
		rightMost = ~(x_or - 1) & x_or;
		i = 0;
		while (i < n)
		{
			if ((arr[i] & rightMost) == 0)
			{
				// When element is repeated in array
				repeated = repeated ^ arr[i];
			}
			else
			{
				// When element is missing in array
				missing = missing ^ arr[i];
			}
			i += 1;
		}
		i = 1;
		while (i <= n)
		{
			if ((i & rightMost) == 0)
			{
				// When element is repeated in range (1...n)
				repeated = repeated ^ i;
			}
			else
			{
				// When element is missing in range (1...n)
				missing = missing ^ i;
			}
			i += 1;
		}
		self.printArray(arr, n);
		// Display calculated result
		print("\n Missing : ", missing);
		print(" Repeated : ", repeated);
	}
}
func main()
{
	let task: Finding = Finding();
	let arr1: [Int] = [1, 8, 5, 7, 8, 2, 6, 3];
	// Get the size of array arr1
	var n: Int = arr1.count;
	task.repeatingAndMissing(arr1, n);
	let arr2: [Int] = [1, 4, 3, 3];
	// Get the size of array arr2
	n = arr2.count;
	task.repeatingAndMissing(arr2, n);
}
main();

input

 Array Element  1  8  5  7  8  2  6  3
 Missing :  4
 Repeated :  8

 Array Element  1  4  3  3
 Missing :  3
 Repeated :  2
/*
  Kotlin Program for
  Find the repeating and the missing element using xor
*/
class Finding
{
	// Print the given array elements
	fun printArray(arr: Array < Int > , n: Int): Unit
	{
		print("\n Array Element");
		var i: Int = 0;
		while (i < n)
		{
			print(" " + arr[i]);
			i += 1;
		}
	}
	fun repeatingAndMissing(arr: Array < Int > , n: Int): Unit
	{
		if (n < 2)
		{
			return;
		}
		// Define auxiliary variables
		var repeated: Int = 0;
		var missing: Int = 0;
		var i: Int = 0;
		var x_or: Int = arr[i];
		var rightMost: Int;
		i = 1;
		while (i < n)
		{
			x_or = x_or xor arr[i];
			i += 1;
		}
		i = 1;
		while (i <= n)
		{
			x_or = x_or xor 1;
			i += 1;
		}
		// Find rightmost set bits
		rightMost = (x_or - 1).inv() and x_or;
		i = 0;
		while (i < n)
		{
			if ((arr[i] and rightMost) == 0)
			{
				// When element is repeated in array
				repeated = repeated xor arr[i];
			}
			else
			{
				// When element is missing in array
				missing = missing xor arr[i];
			}
			i += 1;
		}
		i = 1;
		while (i <= n)
		{
			if ((i and rightMost) == 0)
			{
				// When element is repeated in range (1...n)
				repeated = repeated xor i;
			}
			else
			{
				// When element is missing in range (1...n)
				missing = missing xor i;
			}
			i += 1;
		}
		this.printArray(arr, n);
		// Display calculated result
		println("\n Missing : " + missing);
		println(" Repeated : " + repeated);
	}
}
fun main(args: Array < String > ): Unit
{
	val task: Finding = Finding();
	val arr1: Array < Int > = arrayOf(1, 8, 5, 7, 8, 2, 6, 3);
	// Get the size of array arr1
	var n: Int = arr1.count();
	task.repeatingAndMissing(arr1, n);
	val arr2: Array < Int > = arrayOf(1, 4, 3, 3);
	// Get the size of array arr2
	n = arr2.count();
	task.repeatingAndMissing(arr2, n);
}

input

 Array Element 1 8 5 7 8 2 6 3
 Missing : 4
 Repeated : 8

 Array Element 1 4 3 3
 Missing : 3
 Repeated : 2


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