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;
bool visit[10000006];
int least_prime[10000006];
long long int mark[1000005];
long long int mp[1000005];
long long int br[1000005];
vector<int> primes;
bool prime[10000006];
void sieve() {
  memset(prime, true, sizeof(prime));
  prime[0] = prime[1] = false;
  for (long long int i = 2; i * i <= 10000006; i++) {
    if (prime[i]) {
      for (long long int j = i * 2; j <= 10000006; j += i) {
        prime[j] = false;
      }
    }
  }
}
void pre() {
  int tot = 10000003;
  least_prime[1] = 1;
  for (long long int i = 2; i < tot; i += 2) least_prime[i] = 2;
  for (long long int i = 3; i < tot; i += 2) {
    if (!visit[i]) {
      least_prime[i] = i;
      for (long long int j = i; (j * i) < tot; j += 2) {
        if (!visit[j * i]) {
          visit[j * i] = true;
          least_prime[j * i] = i;
        }
      }
    }
  }
}
int main() {
  long long int n, x, i, ans, p;
  cin >> n >> x;
  for (i = 0; i < n; i++) {
    cin >> br[i];
    mp[br[i]]++;
  }
  ans = 0;
  int like = 0;
  for (i = 0; i < n; i++) {
    if (!mark[br[i]]) {
      p = br[i] ^ x;
      if (p == br[i])
        ans += mp[br[i]] * (mp[br[i]] - 1) / 2;
      else
        ans += mp[br[i]] * mp[p];
      mark[br[i]] = mark[p] = 1;
    }
  }
  cout << ans << "\n";
  return 0;
}
