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>.
public class Problem {

    private static int[] arr = new int[100001];

    public static void main(String[] args) {
        java.util.Scanner input = new java.util.Scanner(System.in);
        int n = input.nextInt();
        int x = input.nextInt();
        for (int i = 0; i < n; i++) {
            arr[input.nextInt()]++;
        }
        long sum = 0;
        if (x > 0) {
            for (int i = 0; i < arr.length; i++) {
                if (arr[i] == 0) {
                    continue;
                }
                if ((i ^ x) < arr.length) {
                    sum += ((long)arr[i ^ x] *(long)arr[i]);
                    arr[i ^ x] = 0;
                }
            }
        } else {
            for (int i = 0; i < arr.length; i++) {
                if (arr[i] == 0) 
                    continue;
                sum += (Math.multiplyExact((long)arr[i] ,(long)(arr[i]-1)) / (long)2);
            }
        }
        System.out.print(sum);
    }
}
