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.PrintWriter;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Iterator;
import java.util.Set;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.Random;
import java.util.StringTokenizer;

public class CF_B {
	public static void main(String[] args) {
		new CF_B().run();
	}
	public void run(){
		FastScanner sc = new FastScanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		
		int n = sc.nextInt();
		int x = sc.nextInt();
		
		HashMap<Integer,Integer> map = new HashMap<Integer,Integer>();
		for(int i = 0;i<n;i++){
			int tmp = sc.nextInt();
			if(map.containsKey(tmp)){
				int times = map.get(tmp);
				map.put(tmp, times+1);
			}else{
				map.put(tmp, 1);
			}
		}
		long pairs = 0;
		Set<Integer> set = map.keySet();
		Set<Integer> hasCal = new HashSet<Integer>();
		Iterator<Integer> it = set.iterator();
		Iterator<Integer> it1 = set.iterator();
		//0异或任何数都为原来的数
		if(x == 0){
			while(it1.hasNext()){
				int tmp = it1.next();
				//如果50000个1和50000个2，x是0的话，结果是2500000000，越界，所以用long
				if(map.containsKey(tmp) && map.get(tmp)>=2)
					pairs+=1L*map.get(tmp)*(map.get(tmp)-1)/2;
			}
			out.println(pairs);
			out.close();
			return ;
		}
		while(it.hasNext()){
			int one = it.next();
			if(hasCal.contains(one))
				continue;
			int another = x^one;
			//50000*50000会越界
			if(map.containsKey(another)){
				pairs+=1L*map.get(one)*map.get(another);
				hasCal.add(another);
				
			}
		}
		out.println(pairs);
		out.close();
	}
}
class FastScanner {
	public BufferedReader br;
	public StringTokenizer st;
	
	public FastScanner(InputStream in){
		br = new BufferedReader(new InputStreamReader(in));
		st = null;
	}
	
	public boolean hasMoreTokens(){
		while(st == null || !st.hasMoreTokens()){
			String line = null;
			try {
				line = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			if(line == null)
				return false;
			st = new StringTokenizer(line);
		}
		return true;
	}
	
	public String next(){
		if(!hasMoreTokens())
			return null;
		return st.nextToken();
	}
	
	public int nextInt(){
		return Integer.parseInt(next());
	}
	
	public long nextLong(){
		return Long.parseLong(next());
	}
	
	public int[] readIntArray(int n){
		int[] array = new int[n];
		for(int i = 0;i<n;i++){
			array[i] = nextInt();
		}
		return array;
	}
	
	public long[] readLongArray(int n){
		long[] array = new long[n];
		for(int i = 0;i<n;i++){
			array[i] = nextLong();
		}
		return array;
	}
	
	public String[] readStringArray(int n){
		String[] a = new String[n];
		for(int i = 0;i<n;i++){
			a[i] = next();
		}
		return a;
	}
	public void shuffleIntArray(int[] a){
		Random rand = new Random();
		for(int i = 0;i<a.length;i++){
			int x = rand.nextInt(i+1);  //0~i
			int t = a[i];
			a[i] = a[x];
			a[x] = t;
		}
	}
	
	//只适合比较小的数，太大的数会越界
	public long factorial(long n){
		if(n==0)
			return 1L;
		long res = 1;
		for(long i = 2;i<=n;i++){
			res*=i;
		}
		return res;
	}
	
	//只适合比较小的数，太大的数会越界
	public long combination(long mom,long son){
		return factorial(mom)/(factorial(son)*factorial(mom-son));
	}
}
