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.HashMap;
import java.util.Scanner;

public class one {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner s = new Scanner(System.in);
		int n = s.nextInt();
		int x = s.nextInt();
		int[] array = new int[n];
		HashMap<Integer,Long> map = new HashMap<>();
		for(int i=0;i<n;i++){
			array[i] = s.nextInt();
			if(map.containsKey(array[i])){
				map.put(array[i], map.get(array[i])+1);
			}else{
				map.put(array[i], (long) 1);
			}
		}
		long count=0;
		for(int i:map.keySet()){
			if((i^x)==i){
				long temp = map.get(i);
				count += (temp*(temp-1));
				continue;
			}
			if(map.containsKey(i^x)){
				count+= map.get(i)*map.get(i^x);
			}
		}
		
		System.out.println(count/2);
	}
}