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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.StringTokenizer;

public class Round381 
{

	 
	public static void main(String[]args)throws Throwable
	{
		SC sc = new SC(System.in);
		int n = sc.nextInt(); int x = sc.nextInt();
		int arr[] = new int[n];
		int cnt[] = new int[100000+10];
		for(int i = 0 ; i < n ; ++i)
		{
			arr[i] = sc.nextInt();
			cnt[arr[i]]++;
		}
		long ans = 0;
		for(int i = 0 ; i < n ; ++i)
		{
			int need = x ^ arr[i];
			if(need >= 100001)
				continue;
			ans += cnt[need];
			if(arr[i] == need)
				--ans;
		}
		System.out.println(ans / 2L);
		
	}
	
	static class SC
	{
		BufferedReader br;
		StringTokenizer st;

		SC(InputStream s)
		{
			br = new BufferedReader(new InputStreamReader(s));
		}

		String next() throws IOException
		{
			while(st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		int nextInt() throws NumberFormatException, IOException
		{
			return Integer.parseInt(next());
		}

		long nextLong() throws NumberFormatException, IOException
		{
			return Long.parseLong(next());
		}
	}
}
