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 Main {
  
  public int getThird(int value, int x){
    //we have to count until 17 bits
    return value ^ x;
  }
  
  public long getPairs(int[] ar, int x){
    long ans = 0;
    for(int i = 0; i < ar.length; i++){
      if(ar[i] > 0){
        int val = getThird(i, x);
//        System.out.println(i + " " + val+ " "+ar[i]+" "+ar[val]);
        if(val == i){
          ans += (long) ar[val] * ((long) ar[i] - 1);
        }
        else{
          ans += (long) ar[val] * (long) ar[i];
        }
        
      }
    }
    return ans / 2L;
  }
  
  public static void main(String[] args) {
    Scanner cin = new Scanner(System.in);
    Main sol = new Main();
    while(cin.hasNextInt()){
      int n = cin.nextInt();
      int x = cin.nextInt();
      int[] ar = new int[131072];
      for(int i = 0; i < n; i++){
        int value = cin.nextInt();
        ar[value]++;
      }
      System.out.println(sol.getPairs(ar, x));
    }
  }
}