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.*;
import java.io.*;

public class Main {
	BufferedReader in;
	PrintWriter out;
	StringTokenizer str;

	private String next() throws Exception {
		if (str == null || !str.hasMoreElements())
			str = new StringTokenizer(in.readLine());
		return str.nextToken();
	}

	private int nextInt() throws Exception {
		return Integer.parseInt(next());
	}
	
	public void run() throws Exception {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		int N = nextInt(), x = nextInt();
		Map<Integer, Integer> map = new HashMap<>();
		for(int i = 0; i < N; ++i) {
			int n = nextInt();
			if (!map.containsKey(n)) {
				map.put(n, 0);
			}
			map.put(n, map.get(n) + 1);
		}
		long ret = 0;
		for(int i : map.keySet()) {
			int lookup = i ^ x;
			if (i > lookup) continue;
			if (map.containsKey(lookup)) {
				if (i == lookup) {
					int l = map.get(i);
					ret += 1L * l * (l - 1) / 2;
				} else {
					ret += 1L * map.get(lookup) * map.get(i);	
				}
			}
		}
		out.println(ret);
		out.close();
	}

	public static void main(String args[]) throws Exception {
		new Main().run();
	}
}
