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 s, c, i, k, j, n, m, z, b;
long long t, cas, d, i1, a, i2, r, w;
long long ta[1000000];
vector<pair<int, int> > vk2;
string we1, we2, we;
vector<long long> vz, vk, va;
map<string, long long> mp;
map<string, long long>::iterator it;
int main() {
  cin >> n >> b;
  vk.assign(100002, 0);
  va.assign(100002, 0);
  k = 0;
  for (i = 0; i < n; i++) {
    cin >> a;
    vk[a]++;
    k = max(k, a);
  }
  z = 0;
  if (b > 0) {
    for (i = 1; i <= k; i++) {
      d = i ^ b;
      if (d <= k) z += vk[i] * vk[d];
    }
  } else {
    for (i = 1; i <= k; i++) {
      z += (vk[i] * (vk[i] - 1)) / 2;
    }
    z *= 2;
  }
  cout << z / 2 << endl;
  return 0;
}
