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.File;
import java.util.HashMap;
import java.util.Scanner;
import java.util.StringTokenizer;

public class p025 {
	public static void main(String args[]) throws Exception {
// 		StringTokenizer stok = new StringTokenizer(new Scanner(new File("C:/Users/Arunkumar/Downloads/input.txt")).useDelimiter("\\A").next());
		StringTokenizer stok = new StringTokenizer(new Scanner(System.in).useDelimiter("\\A").next());
		StringBuilder sb = new StringBuilder();
		int n = Integer.parseInt(stok.nextToken());
		int x = Integer.parseInt(stok.nextToken());
		HashMap<Long,Long> mp = new HashMap<Long,Long>();
		add(mp,Long.parseLong(stok.nextToken()));
		long tot = 0;
		for(int i=1;i<n;i++) {
			long v = Long.parseLong(stok.nextToken());
			long u = v^x;
			tot += get(mp,u);
			add(mp,v);
		}
		System.out.println(tot);
	}
	
	private static long get(HashMap<Long, Long> mp, long k) {
		if(mp.get(k)==null) return 0;
		return mp.get(k);
	}

	private static void add(HashMap<Long, Long> mp, long k) {
		Long v = mp.get(k);
		if(v==null) mp.put(k, 1l);else mp.put(k, v.longValue()+1);
	}
}