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>.
import java.io.*;
import java.math.BigInteger;
import java.util.*;

public class A558 {
  static BufferedReader in = null;
  static PrintWriter out = null;
  static StringTokenizer st = new StringTokenizer("");

  public static void main(String[] args) {
    try {
      in = new BufferedReader(new InputStreamReader(System.in));
      out = new PrintWriter(System.out);
      solve();
      out.close();
    } catch (Exception e) {
      e.printStackTrace();
    }
  }

  public static String readString() {
    while (!st.hasMoreTokens()) {
      try {
        st = new StringTokenizer(in.readLine(), " \n\r\t:");
      } catch (Exception e) {
        e.printStackTrace();
      }
    }
    return st.nextToken();
  }


  public static int readInt() {
    return Integer.parseInt(readString());
  }

  public static long readLong() {
    return Long.parseLong(readString());
  }

  public static int MAX = 100001;

  private static void solve() throws IOException {
    int n = readInt();
    int[] a = new int[n];
    int x = readInt();
    long ans = 0;
    long[] counts = new long[MAX];
    for (int i = 0; i < n; i++) {
      a[i] = readInt();
      counts[a[i]]++;
    }
    if(x == 0){
      for (int i = 0; i < MAX; i++) {
        ans+=(counts[i] * (counts[i] - 1)/2);
      }
      out.print(ans);
      return;
    }
    for (int i = 0; i < MAX; i++) {
      int myPair = x ^ i;
      if (myPair >= 0 && myPair < MAX) {
        ans += counts[i] * counts[myPair];
      }
    }
    out.print(ans/2);

  }

  private static long lcm(long l, long i) {
    return l / gcd(l, i) * i;
  }

  private static long gcd(long l, long i) {
    while (l > 0 && i > 0) {
      if (l > i)
        l %= i;
      else
        i %= l;
    }
    return l + i;
  }


}




