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.*;
import java.io.*;

public class Solution
{
	public static void main(String[] args) throws IOException
	{
		Scanner s = new Scanner(System.in);
		BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out));
		int n = s.nextInt();
		long x = s.nextLong();
		
		int[] array = new int[n];
		long[] maximum = new long[100001];
		
		for(int i=0;i<n;i++)
		{
			array[i]=s.nextInt();
			maximum[array[i]]++;
		}
		
		//Arrays.sort(array);
		//LinkedList<>
		long sum=0;
		
		
		for(int i=0;i<100001;i++)
		{
			if(maximum[i]>0)
			{
				long d = (long)i^x;
				//System.out.println(d);
				if(d<=100000)
				{
					int p=(int)d;
				//	System.out.println(p+" "+i);
				if((i==p && maximum[i]>=2))
				{
					sum+=(maximum[p]*(maximum[p]-1)*1l)/2;
					//System.out.println("Yo");
				}
				else if(i!=p && p<=100000)
				{	sum+=maximum[i]*maximum[p]*1l;
					maximum[p]=0;
					maximum[i]=0;
				}
				}
			}
			
		}
		//;
		//sum*=1l;
		out.write(Long.toString(sum));
		out.flush();
		out.close();
		
	}
}