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;
long long int a[100009], cnt[100009];
int bin_search(int beg, int end, long long int val) {
  int mid = (beg + end) / 2;
  while (beg <= end) {
    mid = (beg + end) / 2;
    if (a[mid] == val) {
      return mid;
    } else if (a[mid] < val) {
      beg = mid + 1;
    } else if (a[mid] > val) {
      end = mid - 1;
    }
  }
  return -1;
}
int main() {
  long long int n, x, k;
  cin >> n >> x;
  for (int i = (0); i <= (n - 1); i++) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  sort(a, a + n);
  long long int sum = 0, j;
  if (x == 0) {
    for (int i = 0; i <= n - 2; i++) {
      sum += ((cnt[a[i]] - 1) * (cnt[a[i]])) / 2;
      while ((i < n - 2) && a[i] == a[i + 1]) {
        i++;
      }
    }
  } else {
    for (int i = 0; i <= n - 2; i++) {
      k = a[i] ^ x;
      j = bin_search(i + 1, n - 1, k);
      if (j != -1) {
        sum += cnt[a[j]] * cnt[a[i]];
      }
      while ((i < n - 2) && a[i] == a[i + 1]) {
        i++;
      }
    }
  }
  cout << sum << endl;
  return 0;
}
