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.StringTokenizer;

public class R383D2P2 {

	public static void main(String[] args) throws Exception {
		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));

		StringTokenizer st = new StringTokenizer(bf.readLine());
		int n = Integer.parseInt(st.nextToken());
		int x = Integer.parseInt(st.nextToken());
		st = new StringTokenizer(bf.readLine());
		long[] freqs = new long[100001];
		for (int i = 0; i < n; i++) {
			int val = Integer.parseInt(st.nextToken());
			freqs[val]++;
		}

		long numWays = 0;
		for (int i = 0; i < freqs.length; i++) {
			if (freqs[i] > 0) {
				int diff = x ^ i;
				if (diff < freqs.length) {
					if (diff == i) { // x == 0
						numWays += freqs[i] * (freqs[i]-1);
					} else {
						numWays += freqs[i] * freqs[diff];
					}
				}
			}
		}
		numWays /= 2;
		System.out.println(numWays);
	}

}