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.*;
import java.util.*;
public class div383B
{
	BufferedReader in;
	PrintWriter ob;
	StringTokenizer st;
	public static void main(String[] args) throws IOException {
		new div383B().run();
	}
	void run() throws IOException {
		//in=new BufferedReader(new FileReader("output.txt"));
		in=new BufferedReader(new InputStreamReader(System.in));
		ob=new PrintWriter(System.out);
		solve();
		ob.flush();
	}
	@SuppressWarnings("unchecked")
	void solve() throws IOException {
		int n=ni();
		int x=ni();
		int a[]=new int[n+1];
		int f[]=new int[100001];
		for(int i=1 ; i<=n ;i++) {
			a[i]=ni();
			f[a[i]]++;
		}	
		long ans=0;
		for(int i=1 ; i<=n ; i++) {
			int find = a[i]^x;
			--f[a[i]];
			if(find>0 && find<=100000)
				ans += f[find]; 
		}
		ob.println(ans);
	}
	char nc() throws IOException {
		return ns().charAt(0);
	}
	String ns() throws IOException {
		return nextToken();
	}
	long nl() throws IOException {
		return Long.parseLong(nextToken());
	}
	int ni() throws IOException {
		return Integer.parseInt(nextToken());
	}
	double nd() throws IOException {
		return Double.parseDouble(nextToken());
	} 
	String nextToken() throws IOException {
		if(st==null || !st.hasMoreTokens())
			st=new StringTokenizer(in.readLine());
		return st.nextToken();
	}
	int[] nia(int start,int b) throws IOException {
		int a[]=new int[b];
		for(int i=start;i<b;i++)
			a[i]=ni();
		return a;
	}
	long[] nla(int start,int n) throws IOException {
		long a[]=new long[n];
		for (int i=start; i<n ;i++ ) {
			a[i]=nl();
		}
		return a;
	}
}