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.io.*;
import java.util.*;
public class Solution {
	static Scanner sc=new Scanner(System.in);
	static PrintWriter out=new PrintWriter(System.out);
	//Main Method
	public static void main(String args[]) {
		int test=1;
		while(test-->0) {
			int n=sc.nextInt(),x=sc.nextInt();
			HashMap<Integer,Integer> map=new HashMap<>();
			for(int i=0;i<n;i++) map.merge(sc.nextInt(), 1, Integer :: sum);
			long ans=0;
			for(Integer k: map.keySet()) {
				long v=0;
				if(x==0) {
					v=map.get(k);
					ans+=(v*(v-1))/2;
				}
				else {
					if(map.get(k^x)!=null) {
						v=map.get(k^x);
						map.put(k^x,0);
					}
					ans+=v*map.get(k);
				}
			}
			out.println(ans);
		}
		out.flush();
		out.close();
	}
}
