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 B {

	public static void main(String[] args) throws Exception{
		BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
		
		StringTokenizer st = new StringTokenizer(f.readLine());
		
		int n = Integer.parseInt(st.nextToken());
		int q = Integer.parseInt(st.nextToken());
		int[] ar = new int[n];
		int[] pair = new int[n];
		TreeMap<Integer,Integer> tm = new TreeMap<Integer,Integer>();
		st = new StringTokenizer(f.readLine());
		for(int i = 0;i<n;i++){
			ar[i] = Integer.parseInt(st.nextToken());
			if(!tm.containsKey(ar[i])) tm.put(ar[i], 1);
			else tm.put(ar[i], tm.get(ar[i])+1);
		}
		int logq;
		int tempq = q;
		int[] bitsq = new int[20];
		
		for(logq = 0;1<<logq<=tempq;logq++);
		logq--;
		
		for(int j = logq;j>=0;j--){
			if(1<<j <= tempq){
				bitsq[j] = 1;
				tempq -= (1<<j);
			}
		}
		for(int i = 0;i<n;i++){
			int log;
			int temp = ar[i];
			int[] bits = new int[20];
		
			for(log = 0;1<<log<=temp;log++);
			log--;
			
			for(int j = log;j>=0;j--){
				if(1<<j <= temp){
					bits[j] = 1;
					temp -= (1<<j);
				}
			}
			
			for(int j = 0;j<20;j++){
				if(bits[j] == 1 && bitsq[j] == 1);
				else if(bits[j] == 1 && bitsq[j] == 0) pair[i] += 1<<j;
				else if(bits[j] == 0 && bitsq[j] == 1) pair[i] += 1<<j;
			}
		}
		long ans = 0;
		for(int i = 0;i<n;i++){
			if(tm.containsKey(pair[i])){
				ans += tm.get(pair[i]);
				if(pair[i] == ar[i]) ans --;
			}
		}
		System.out.println(ans/2);
	}

}
