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.PrintStream;
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.List;
import java.util.Scanner;

public class CF742B {
	/**
	 * solution result
	 */
	public static class Answer {
		long n;
		public Answer(long n) {
			this.n = n;
		}

		@Override
		public String toString() {
			return Long.toString(n);
		}
	};

	/**
	 * input parser
	 */
	public static class Input {
		int n;
		int x;
		int a[];
		public Input(Scanner sc) {
			n = sc.nextInt();
			x = sc.nextInt();
			a = new int[n];
			for (int i=0; i<n; i++)
				a[i] = sc.nextInt();
		}
	};

	/**
	 * solver function
	 */
	public static Answer solve(Input in) {
		int idx[] = new int[1<<17]; // to allow all 17bits set +1
		for (int i=0; i<in.a.length; i++)
			idx[in.a[i]]++;
		long cnt = 0;
		for (int i=0; i<in.a.length; i++) {
			int xor = in.x^in.a[i];
			idx[in.a[i]]--;
			cnt += idx[xor];
		}
		Answer ans = new Answer(cnt);
		return ans;
	}

	/**
	 * standard CodeForces executable mantra
	 */
	public static void proc(Scanner sc, PrintStream out) {
		Input in = new Input(sc);
		out.println(solve(in));
	}

	public static void main(String[] args) {
		try (Scanner sc = new Scanner(System.in)) {
			proc(sc, System.out);
		}
	}
}
