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.Arrays;
import java.util.Scanner;

public class BitXor3 {

	public static void main(String[] args) 
	{
		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt();
		int x = sc.nextInt();
		int a[] = new int[n];
		long h[] = new long[10000000];
		for(int i=0;i<n;i++)
		{
			a[i] = sc.nextInt();
			long c = h[a[i]];
			h[a[i]] = ++c;
			//System.out.println(h[a[i]]);
		}
		//Arrays.sort(a);
		
		long count=0;
		for(int i=0;i<n;i++)
		{
			int b=-1;
			if(h[a[i]]>0)
				 b = a[i]^x;
			else
				continue;
			//System.out.println("h["+b+"] : "+h[b]);
			if(a[i]==b)
			{
				if(h[b] > 1)
				{
					count = count + (h[b]*(h[b]-1))/2;
					h[a[i]] = 0;
					//System.out.println("count : "+count+" | "+a[i]+" "+b);
				}
			}
			else if(h[b]>0)
			{
				
				count = count+h[a[i]]*h[b];
				h[a[i]] = 0;
				//h[b] = 0;
				//System.out.println(a[i]+" "+b);
			}
			
		}
		System.out.println(count);
	}

}
