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.BufferedReader;
        import java.io.IOException;
        import java.io.InputStreamReader;

/**
 * Created by Shahjahan on 4/29/2017.
 */
public class B {

    public static final int LIMIT = 100005;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] split1 = br.readLine().split(" ");
        int n = Integer.parseInt(split1[0]);
        int x = Integer.parseInt(split1[1]);

        int numbers[] = new int[n + 1];
        boolean mark[] = new boolean[LIMIT];
        int count[] = new int[LIMIT];
        String[] split2 = br.readLine().split(" ");
        for (int i = 0; i < n; i++) {
            int number = Integer.parseInt(split2[i]);
            numbers[i] = number;
            count[number]++;
        }

        long sum = 0;
        if (n != 1) {
            if (x == 0) {
                for (int i = 0; i < n; i++) {
                    int left = numbers[i];
                    if (count[left] > 1) {
                        count[left]--;
                        sum += ((long) count[left]);
                    }
                }
            } else {
                for (int i = 0; i < n; i++) {
                    int left = numbers[i];
                    if (mark[left] == false) {
                        int xor = left ^ x;
                        if (xor < LIMIT) {
                            if (left == xor)
                                sum += (long) count[left] - 1;
                            else
                                sum += ((long) count[left] * (long) count[xor]);
                            mark[left] = true;
                            mark[xor] = true;
                        }
                    }
                }
            }
        }

        System.out.println(sum);
    }
}