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>.
//package longchallenge;

import java.util.ArrayList;
import java.util.Scanner;

public class amrc {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int n;
		
		//System.out.println(1<<2);
		Scanner scnr=new Scanner(System.in);
		 n=scnr.nextInt();
		int x=scnr.nextInt();
		int ar[]=new int[1000000+1];
		int ar1[]=new int[n];
		for(int i=0;i<n;i++)
		{
			int temp=scnr.nextInt();
			ar1[i]=temp;
			ar[temp]++;
		}
		ArrayList<Integer> list=new ArrayList();
		
		for(int i=0;i<20;i++)
		{
			
		if(((1<<i)&x)!=0)
		{
			//System.out.println((i));
			list.add(i);
		}
		}
		//System.out.println(207^208);
	//	System.out.println(list);
		long re=0;
		for(int i=0;i<n;i++)
		{
			
			int temp=ar1[i];
			for(int j=0;j<list.size();j++)
			{
				if((ar1[i]&(1<<list.get(j)))!=0)
				temp=temp-(1<<list.get(j));
				else
					temp+=1<<list.get(j);
			}
			if(ar[ar1[i]]>0)
			{
				ar[ar1[i]]--;
			re+=ar[temp];
			}
		}
		System.out.println(re);
		

	}
	
}

