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;
const int N = 1e5 + 5;
bool done[N];
long long a[N];
map<long long, long long> cnt;
struct UnionFind {
  int sz[N], par[N];
  void initialize(int n) {
    for (int i = 0; i <= n; i++) {
      sz[i] = 1, par[i] = i;
    }
  }
  int root(int u) { return par[u] == u ? u : par[u] = root(par[u]); }
  bool merge(int u, int v) {
    u = root(u), v = root(v);
    if (u == v) return false;
    if (sz[u] > sz[v]) swap(u, v);
    par[v] = u, sz[u] += sz[v];
    return true;
  }
} helper;
int main() {
  helper.initialize(N - 2);
  int n, x;
  cin >> n >> x;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    long long need = a[i] ^ x;
    if (!cnt.count(need)) continue;
    if (a[i] == need and !done[need]) {
      ans += cnt[need] * (cnt[need] - 1) / 2;
      done[need] = 1;
      continue;
    }
    if (!helper.merge(a[i], need)) continue;
    ans += cnt[need] * cnt[a[i]];
  }
  cout << ans << endl;
}
