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 javafx.util.Pair;

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

/**
 * Created by chao on 16/11/13.
 */
public class A {


    public static void main(String[] args) throws IOException {
        PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        String[] nums = br.readLine().split(" ");
        int n = Integer.parseInt(nums[0]);
        int x = Integer.parseInt(nums[1]);
        nums = br.readLine().split(" ");
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(nums[i]);
        }

        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < n; i++) {
            if (map.containsKey(a[i])) {
                map.put(a[i], map.get(a[i])+1);
            } else {
                map.put(a[i], 1);
            }
        }

        long ans = 0;
        for (int i = 0; i < n; i++) {
            if (map.containsKey(a[i]^x)) {
                ans += map.get(a[i]^x);
                if (x == 0) ans--;
            }
        }
        ans /= 2;
        pw.println(ans);


        pw.flush();
        pw.close();
        br.close();

    }

}



