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 MX = 500001;
void task(long long case_no) {
  long long n, x;
  cin >> n >> x;
  vector<long long> a(n);
  map<long long, long long> freq;
  for (auto &i : a) {
    cin >> i;
    ++freq[i];
  }
  map<long long, long long> mark;
  long long ans = 0;
  for (auto i : freq) {
    if (mark[i.first]) continue;
    long long p = i.first ^ x;
    mark[p] = 1;
    if (!x)
      ans += (freq[p] * (freq[p] - 1)) / 2;
    else
      ans += freq[p] * i.second;
  }
  cout << ans << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t = 1, i = 0;
  while (t--) {
    task(++i);
  }
  return 0;
}
