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 long long mod = 1e9 + 7, mod2 = 2e9;
double Pi = 3.141592653589793238;
const double long rad = Pi / 180.0;
template <class tone>
tone gcd(tone a, tone b) {
  return b == 0 ? a : gcd(b, a % b);
}
template <class tone>
tone lcm(tone a, tone b) {
  return (a * b) / gcd(a, b);
}
template <class tone>
tone FP(tone b, tone p) {
  if (p == 0) return 1;
  return (p % 2 == 1) ? (b)*FP(b * b, p / 2) : FP(b * b, p / 2);
}
long long sum[1000005] = {};
long long v[100005] = {};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t, k;
  scanf("%I64d%I64d", &t, &k);
  for (int i = (0); i < (t); i++) {
    scanf("%I64d", &v[i]);
    sum[v[i]]++;
  }
  long long co = 0;
  for (int i = (0); i < (t); i++) {
    unsigned long long z = v[i] ^ k;
    if (z > 100001) continue;
    if (z == v[i]) {
      co += sum[z] * (sum[z] - 1) / 2;
    } else {
      co += (sum[v[i]] * sum[z]);
    }
    sum[v[i]] = 0;
    sum[z] = 0;
  }
  printf("%I64d", co);
}
