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 Spoj;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

public class CHFNFRN {	
	
	
	// boolean array isSpecial
	// id to msg 
	//	id to popularity
	public static void main(String[] args) {
		int max = (int)(1E5)+5 ;
	//	System.out.println(max);
		InputReader ir = new InputReader(System.in);
		int n = ir.nextInt();
		int x = ir.nextInt();
		int [] a = new int[n] ;
		int [] c = new int[max];
		for(int i=0;i<c.length;i++) 
			c[i] = 0;
		long pairs =0 ;
		for(int i=0;i<n;i++) {
			a[i] = ir.nextInt();
			int key = a[i]^x;
			//check if key is there or not 
			if(key<=max-5) {
				pairs+= c[key];
				c[a[i]]++;
			}
		}
		//to do: case of 0  
		System.out.println(pairs);
	}	

 static class InputReader {
	private BufferedReader reader ;
	private StringTokenizer tokenizer;

	public InputReader(InputStream stream) {
		reader = new BufferedReader(new InputStreamReader(stream));
		tokenizer = null;
	}

	public String next() {
		while(tokenizer==null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(reader.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}

	public float nextFloat() {
		return Float.parseFloat(next());
	}

	public String nextLine() {
		try {
			return reader.readLine();
		} catch (IOException e) {
			throw new RuntimeException(e);
		}
	}
}

}