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;
vector<long long int> a;
long long int first(long long int beg, long long int end, long long int x) {
  if (end >= beg) {
    long long int mid = beg + (end - beg) / 2;
    if ((mid == 0 || (x > a[mid - 1])) && (a[mid] == x))
      return mid;
    else if (x > a[mid])
      return first(mid + 1, end, x);
    else
      return first(beg, mid - 1, x);
  }
  return -1;
}
long long int last(long long int beg, long long int end, long long int x) {
  if (end >= beg) {
    long long int mid = beg + (end - beg) / 2;
    if ((mid == (a.size() - 1) || (x < a[mid + 1])) && (a[mid] == x))
      return mid;
    else if (x < a[mid])
      return last(beg, mid - 1, x);
    else
      return last(mid + 1, end, x);
  }
  return -1;
}
long long int no_of_ocur(long long int beg, long long int end,
                         long long int x) {
  long long int ans = 0;
  if (binary_search(a.begin() + beg, a.end(), x)) {
    long long int fs, lst;
    fs = first(beg, end, x);
    lst = last(beg, end, x);
    ans = lst - fs + 1;
  }
  return ans;
}
int main() {
  long long int n;
  long long int x;
  cin >> n >> x;
  a.resize(n);
  long long int i, j;
  for (i = 0; i < n; i++) cin >> a[i];
  sort(a.begin(), a.end());
  long long int ans = 0;
  if (x > 0) {
    for (i = 0; i < (n - 1); i++) {
      long long int tofind = x ^ a[i];
      long long int cur = no_of_ocur(i + 1, n - 1, tofind);
      ans = ans + cur;
    }
  } else {
    long long int cur = a[0];
    long long int count = 1;
    for (i = 1; i < n; i++) {
      if (a[i] == cur)
        count++;
      else {
        cur = a[i];
        ans = ans + count * (count - 1) / 2;
        count = 1;
      }
    }
    ans = ans + count * (count - 1) / 2;
  }
  cout << ans << "\n";
  return 0;
}
