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>.

// ~/BAU/ACM-ICPC/Teams/A++/BlackBurn95

import java.io.*;
import java.util.*;
import java.math.*;
import static java.lang.Math.*;
import static java.lang.Integer.parseInt;
import static java.lang.Long.parseLong;
import static java.lang.Double.parseDouble;
import static java.lang.String.*;

public class Main {

    public static void main(String[] args) throws IOException {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
                                              //(new FileReader("input.in"));
        StringBuilder out = new StringBuilder();
        StringTokenizer tk;
       
        tk = new StringTokenizer(in.readLine());
        int n = parseInt(tk.nextToken()),x = parseInt(tk.nextToken());
        
        tk = new StringTokenizer(in.readLine());
        long [] f = new long[100001];
        Set<Integer> s = new HashSet<>();
        int tmp;
        for(int i=0; i<n; i++) {
            tmp = parseInt(tk.nextToken());
            f[tmp]++;
            s.add(tmp);
        }
        
        long ans = 0;
        
        for(int num : s) {
            tmp = 0;
            for(int i=0; i<30; i++) {
                if((x & (1<<i))!=0) {
                    if((num & (1<<i))==0) tmp |= (1<<i);
                } else 
                    tmp |= (num & (1<<i));
            }
            
            if(tmp > 100000) continue;
            
            if(tmp>num) 
                ans += f[num]*f[tmp];
            else if(tmp==num) ans += f[num]*(f[num]-1)/2;
        }
        
        System.out.println(ans);
    }
}