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.*;
import java.util.*;

public class ProblemB {

	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] in = br.readLine().split(" ");
		int n = Integer.parseInt(in[0]);
		int x = Integer.parseInt(in[1]);
		String[] nums = br.readLine().split(" ");
		int[] num = new int[nums.length];
		
		for(int i=0; i<nums.length; i++) {
			num[i] = Integer.parseInt(nums[i]);
		}
		
		HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
		for(int i=0; i<num.length; i++) {
			if(map.containsKey(num[i])) {
				map.put(num[i], map.get(num[i]) + 1);
			} else {
				map.put(num[i], 1);
			}
		}
		
		long ways = 0;
		for(int i=0; i<num.length; i++) {
			int sol = xorsub(num[i], x);
			if(map.containsKey(sol)) {
				if(sol != num[i])
					ways += map.get(sol);
				else
					ways += map.get(sol) - 1;
			}
		}
		
		System.out.println(ways/2);
	}
	
	public static int xorsub (int a, int x) {
		int selector = 1;
		int out = 0;
		
		for(int i=0; i<32; i++) {
			int abit = a & selector;
			int xbit = x & selector;
			
			if((abit > 0 && xbit == 0) || (abit == 0 && xbit > 0)) {
				out += selector;
			}
			selector = selector << 1;
		}
		
		return out;
	}
}
