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 double EPS = 1e-9;
const int INF = 0x7f7f7f7f;
const double PI = acos(-1.0);
template <class T>
inline T _abs(T n) {
  return ((n) < 0 ? -(n) : (n));
}
template <class T>
inline T _max(T a, T b) {
  return (!((a) < (b)) ? (a) : (b));
}
template <class T>
inline T _min(T a, T b) {
  return (((a) < (b)) ? (a) : (b));
}
template <class T>
inline T _swap(T& a, T& b) {
  a = a ^ b;
  b = a ^ b;
  a = a ^ b;
}
template <class T>
inline T gcd(T a, T b) {
  return (b) == 0 ? (a) : gcd((b), ((a) % (b)));
}
template <class T>
inline T lcm(T a, T b) {
  return ((a) / gcd((a), (b)) * (b));
}
struct debugger {
  template <typename T>
  debugger& operator,(const T& v) {
    cerr << v << " ";
    return *this;
  }
} dbg;
int bitOn(int N, int pos) { return N = N | (1 << pos); }
int bitOff(int N, int pos) { return N = N & ~(1 << pos); }
bool bitCheck(int N, int pos) { return (bool)(N & (1 << pos)); }
map<int, int> mp;
vector<int> vec;
int main() {
  int n, x;
  while (~scanf("%d%d", &n, &x)) {
    for (int i = 1; i <= n; i++) {
      int p;
      cin >> p;
      vec.push_back(p);
      mp[p]++;
    }
    long long cc = 0;
    for (int i = 0; i < n; i++) {
      int m = vec[i];
      mp[m]--;
      long long dekhi = m ^ x;
      cc += mp[dekhi];
    }
    cout << cc << endl;
  }
}
