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;
template <class T>
inline T BMOD(T p, T e, T m) {
  T ret = 1;
  while (e) {
    if (e & 1) ret = (ret * p) % m;
    p = (p * p) % m;
    e >>= 1;
  }
  return (T)ret;
}
template <typename T>
inline T POW(T B, T P) {
  if (P == 0) return 1;
  if (P & 1)
    return B * POW(B, P - 1);
  else
    return SQR(POW(B, P / 2));
}
template <typename T>
inline T ModInv(T b, T m) {
  return BMOD(b, m - 2, m);
}
template <typename T>
inline T ABS(T a) {
  if (a < 0)
    return -a;
  else
    return a;
}
template <typename T>
inline T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
template <typename T>
inline T lcm(T a, T b) {
  if (a < 0) return lcm(-a, b);
  if (b < 0) return lcm(a, -b);
  return a * (b / gcd(a, b));
}
int dx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dy[] = {-1, 0, 1, -1, 1, -1, 0, 1};
int n, x, trck[200005], arr[100005];
int main() {
  scanf("%d%d", &n, &x);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
    trck[arr[i]]++;
  }
  long long int res = 0;
  for (int i = 1; i <= 100000; i++) {
    int nm = i ^ x;
    if (trck[nm] && trck[i]) {
      long long int cnt = (long long int)trck[nm] * trck[i];
      if (x == 0) cnt = (long long int)trck[nm] * (trck[i] - 1);
      res += cnt;
    }
  }
  res /= 2;
  printf("%lld", res), puts("");
  return 0;
}
