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.util.*;
import java.io.*;

public final class ProblemB {
    public static void main(String[] args) {
        final Scanner in = new Scanner(System.in);
        final int n = in.nextInt();
        final int x = in.nextInt();

        final int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
        }

        final Map<Integer, Integer> numCount = new HashMap<Integer, Integer>();
        for (final int i: a) {
            final int oldCount = numCount.getOrDefault(i, 0);
            numCount.put(i, oldCount + 1);
        }


        long pairCount = 0L;
        for (final int i: a) {
            final int b = i ^ x;
            final int countB = numCount.getOrDefault(b, 0);
            pairCount += countB;
            if (b == i) {
                pairCount--;
            }
        }
        pairCount /= 2;

        System.out.println(pairCount);
    }

}