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.InputStreamReader;
import java.util.Map;
import java.util.Scanner;
import java.util.TreeMap;

public class B {
	private void work() {
		Scanner sc = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		int n = sc.nextInt();
		int x = sc.nextInt();
		TreeMap<Integer, Long> a = new TreeMap<Integer, Long>();
		for (int i = 0; i < n; i++) {
			int next = sc.nextInt();
			if (!a.containsKey(next))
				a.put(next, 0L);
			a.put(next, a.get(next) + 1L);
		}

		long res = 0;
		for (Map.Entry<Integer, Long> e : a.entrySet()) {
			long y = e.getValue();
			if (x == 0) {
				res += y * (y - 1);
			} else {
				int z = e.getKey() ^ x;
				if (a.containsKey(z))
					res += y * a.get(z);
			}
		}

		System.out.println(res / 2);

		sc.close();
	}

	public static void main(String[] args) {
		new B().work();
	}

}
