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.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {
	public static void main(String[] args) {
		FastReader in = new FastReader ();
		int n = in.nextInt (), x = in.nextInt (), contadores[] = new int[(int) (2e5 + 20)], a;
		long ans = 0;
		
		for (int i=0; i<n; i++) {
			a = in.nextInt ();
			ans += contadores[a^x];
			contadores[a]++;
		}
	   System.out.println (ans);
	}
	
	static class FastReader {
	    BufferedReader br;
	    StringTokenizer st;

	    public FastReader() {
	        br = new BufferedReader(new InputStreamReader(System.in));
	    }

	    String next() {
	 	   String  line;
	       while (st == null || !st.hasMoreElements()) {
	          line = nextLine ();
	          if (line == null)
	             return null;
	          st = new StringTokenizer(line);
	       }
	       return st.nextToken();
	    }

	    String nextLine() {
	       String str = null;
	       try {
	           str = br.readLine();
	       }
	       catch (IOException e)
	       {}
	       return str;
	    }
	    
	    Integer nextInt() {
	  	   String element = next ();
	       return element==null? null:Integer.parseInt(element);
	    }

	    Long nextLong() {
	  	  String element = next ();
	      return element==null? null:Long.parseLong(element);
	    }

	    Double nextDouble() {
	  	  String element = next ();
	      return element==null? null:Double.parseDouble(element);
	    }
	}
}
