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>.
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int n, x;
  cin >> n >> x;
  long long int a[n];
  long long int i, fl[100005] = {0};
  for (i = 0; i < n; i++) {
    cin >> a[i];
    fl[a[i]]++;
  }
  long long int f[100] = {0};
  long long int k = 0;
  while (x != 0) {
    if (x % 2 == 0) {
      f[k] = 0;
      x = x / 2;
      k++;
    } else {
      f[k] = 1;
      x = x / 2;
      k++;
    }
  }
  long long int z;
  long long int j;
  long long int ans = 0;
  for (i = 0; i < n; i++) {
    z = a[i];
    k = 0;
    long long int f1[100] = {0}, f2[100] = {0};
    while (z != 0) {
      if (z % 2 == 0)
        f1[k] = 0;
      else
        f1[k] = 1;
      k++;
      z = z / 2;
    }
    for (j = 0; j <= 20; j++) {
      if (f[j] == 0 && f1[j] == 1)
        f2[j] = 1;
      else if (f[j] == 1 && f1[j] == 0)
        f2[j] = 1;
    }
    long long int s = 0;
    for (j = 0; j <= 20; j++) {
      if (f2[j] == 1) s = s + pow(2, j);
    }
    if (s > 100000) continue;
    if (a[i] == s) {
      ans = ans + (fl[s] * (fl[s] - 1)) / 2;
    } else
      ans = ans + fl[a[i]] * fl[s];
    fl[a[i]] = 0;
    fl[s] = 0;
  }
  cout << ans << endl;
}
