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.*;

public class ProblemB {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String line1 = in.nextLine();
        String line2 = in.nextLine();

        int n = Integer.parseInt(line1.split(" ")[0]);
        int x = Integer.parseInt(line1.split(" ")[1]);

        String[] numbers = line2.split(" ");

        ArrayList<Integer> arr = new ArrayList<Integer>();
        for(int i = 0; i < numbers.length; i++) {
            arr.add(Integer.parseInt(numbers[i]));
        }

        long count = 0;
        int number = 0;

        Map<Integer, Integer> map = new HashMap<>();

        for (int i = 0; i < arr.size(); i++) {
            if(map.get(arr.get(i)) == null) {
                map.put(arr.get(i), 1);
            }
            else {
                map.put(arr.get(i), map.get(arr.get(i)) + 1);
            }
        }

        for (int i = 0; i < arr.size(); i++) {
            number = arr.get(i) ^ x;
            count += map.get(number) == null ? 0 : map.get(number);
            if(number == arr.get(i)) {
                count--;
            }
        }
        System.out.println(count / 2);
    }

}