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.*;
import java.lang.*;
import java.util.*;
public class Main {
    public static void main(String[] args) throws IOException {
        StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        Scanner sc = new Scanner(System.in);
        Map<Long, Long> mp = new HashMap<>();
        int maxn = (int) 600000;
        long a[] = new long[maxn];
        int n = sc.nextInt();
        long x = sc.nextLong();
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextLong();
            if (mp.containsKey(a[i])) {
                mp.put(a[i], mp.get(a[i]) + 1);
            } else {
                mp.put(a[i], 1L);
            }
        }
        long ans = 0;
        for (int i = 0; i < n; ++i) {
            long cnt = x ^ a[i];
            long add = 0;
            if (mp.containsKey(cnt))
                add = mp.get(cnt);
            if (cnt != a[i])
                ans += add;
            else
                ans += add - 1;
        }
        System.out.print(ans/2);
    }
}