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.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;
public class B {
	public static void main(String[] args)throws Throwable {
		MyScanner sc=new MyScanner();
		PrintWriter pw=new PrintWriter(System.out,true);
		
		int n=sc.nextInt();
		int x=sc.nextInt();
		
		int []a=new int [n];
		int [] other=new int [n];
		TreeMap<Integer,Integer> map=new TreeMap<Integer, Integer>();
		
		for(int i=0;i<n;i++){
			a[i]=sc.nextInt();
			int o=(x ^ a[i]);
			other[i]=o;
			
			if(map.containsKey(a[i]))
				map.put(a[i], map.get(a[i])+1);
			else
				map.put(a[i], 1);
		}
		long cnt=0;
		for(int i=0;i<n;i++){
			if(map.containsKey(a[i])){
				int rem=map.get(a[i])-1;
				if(rem!=0)
					map.put(a[i], rem);
				else
					map.remove(a[i]);
			}
			if(map.containsKey(other[i]))
			cnt+=map.get(other[i]);
		}
		
		
		System.out.println(cnt);
		
		pw.flush();
		pw.close();
	}
	
	static class MyScanner {
	      BufferedReader br;
	      StringTokenizer st;
	 
	      public MyScanner() {
	         br = new BufferedReader(new InputStreamReader(System.in));
	      }
	 
	      String next() {
	          while (st == null || !st.hasMoreElements()) {
	              try {
	                  st = new StringTokenizer(br.readLine());
	              } catch (IOException e) {
	                  e.printStackTrace();
	              }
	          }
	          return st.nextToken();
	      }
	 
	      int nextInt() {
	          return Integer.parseInt(next());
	      }
	 
	      long nextLong() {
	          return Long.parseLong(next());
	      }
	 
	      double nextDouble() {
	          return Double.parseDouble(next());
	      }
	 
	      String nextLine(){
	          String str = "";
		  try {
		     str = br.readLine();
		  } catch (IOException e) {
		     e.printStackTrace();
		  }
		  return str;
	      }

	   }
}
