There are some beautiful girls in Arpa’s land as mentioned before.

Once Arpa came up with an obvious problem:

Given an array and a number x, count the number of pairs of indices i, j (1 ≤ i < j ≤ n) such that <image>, where <image> is bitwise xor operation (see notes for explanation).

<image>

Immediately, Mehrdad discovered a terrible solution that nobody trusted. Now Arpa needs your help to implement the solution to that problem.

Input

First line contains two integers n and x (1 ≤ n ≤ 105, 0 ≤ x ≤ 105) — the number of elements in the array and the integer x.

Second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 105) — the elements of the array.

Output

Print a single integer: the answer to the problem.

Examples

Input

2 3
1 2


Output

1

Input

6 1
5 1 2 3 4 1


Output

2

Note

In the first sample there is only one pair of i = 1 and j = 2. <image> so the answer is 1.

In the second sample the only two pairs are i = 3, j = 4 (since <image>) and i = 1, j = 5 (since <image>).

A bitwise xor takes two bit integers of equal length and performs the logical xor operation on each pair of corresponding bits. The result in each position is 1 if only the first bit is 1 or only the second bit is 1, but will be 0 if both are 0 or both are 1. You can read more about bitwise xor operation here: <https://en.wikipedia.org/wiki/Bitwise_operation#XOR>.
import java.util.*;
import java.math.*;
public class threeEightyThreeB{
	public static BigInteger factorial(long in){
		BigInteger output = new BigInteger("1");
		output = output.multiply( new BigInteger( String.valueOf(in) ) );
		output = output.multiply( new BigInteger( String.valueOf(in - 1) ) ) ;
		output = output.divide( new BigInteger( String.valueOf(2) ) );
		return output;
	}
	public static void main(String args[]){
		Scanner input = new Scanner(System.in);
		long n = input.nextLong();
		long x = input.nextLong();

		input.nextLine();

		Hashtable h = new Hashtable();
		Hashtable h_rep = new Hashtable();
		long i = 0;
		long output = 0;
		while( i < n ){
			long temp1 = input.nextLong(  );

			if( temp1 == x ){
				i++;
				continue;
			}

			if( h_rep.containsKey( temp1 ) ){
				h_rep.put( temp1, (long) ( h_rep.get( temp1 ) ) + 1 );
			} else {
				h_rep.put( temp1, (long)1 );
			}

			i++;
		}

		Enumeration keys = h_rep.keys();
		BigInteger out_b = new BigInteger("0");
      	while(keys.hasMoreElements()) {
            long k  = (long)keys.nextElement();
	  		if( x == 0  ){
	  			if( (long)h_rep.get( k ) > 1 ){
	  				out_b = out_b.add( factorial( (long)h_rep.get( k ) ) );
	  			}
      		}else{
	        	long temp = k ^ x;
	        	if( h_rep.containsKey( temp ) ){
	        		output += (long)h_rep.get( k ) * (long)h_rep.get( temp );
	        		h_rep.remove( k );
	        		h_rep.remove( temp );
	        	}
      		}

      	}
      	if(x == 0){
			System.out.print(out_b);
			return;
      	}
		System.out.print(output);
	}
}