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.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.util.Scanner;



public class D {
public static void main(String args[])
{
	InputReader in=new InputReader(System.in);
	PrintWriter w=new PrintWriter(System.out);
	int n=in.nextInt();
	int x=in.nextInt();
	int freq[]=new int[100000+1];
	int a[]=new int[n];
	for(int i=0;i<n;i++)
	{
		int temp=in.nextInt();
		a[i]=temp;
		freq[temp]++;
	}
	
	if(x!=0){
		long ans=0;
	for(int i=0;i<n;i++)
	{
		if((a[i]^x)<=100000)
			ans+=freq[(a[i]^x)];
	}
	System.out.println(ans/2);
	}
	else
	{
		long ans=0;
		int visited[]=new int[100000+1];
		for(int i=0;i<n;i++)
		{
			if(freq[a[i]^x]>1)
			{
				ans+=(((freq[(a[i]^x)]-1))-visited[a[i]]);
				visited[a[i]]++;
			}
		}
		System.out.println(ans);
	}
	
	
	
}
static class InputReader {
	 
	private final InputStream stream;
	private final byte[] buf = new byte[8192];
	private int curChar, snumChars;
	private SpaceCharFilter filter;

	public InputReader(InputStream stream) {
		this.stream = stream;
	}

	public int snext() {
		if (snumChars == -1)
			throw new InputMismatchException();
		if (curChar >= snumChars) {
			curChar = 0;
			try {
				snumChars = stream.read(buf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (snumChars <= 0)
				return -1;
		}
		return buf[curChar++];
	}

	public int nextInt() {
		int c = snext();
		while (isSpaceChar(c))
			c = snext();
		int sgn = 1;
		if (c == '-') {
			sgn = -1;
			c = snext();
		}
		int res = 0;
		do {
			if (c < '0' || c > '9')
				throw new InputMismatchException();
			res *= 10;
			res += c - '0';
			c = snext();
		} while (!isSpaceChar(c));
		return res * sgn;
	}

	public long nextLong() {
		int c = snext();
		while (isSpaceChar(c))
			c = snext();
		int sgn = 1;
		if (c == '-') {
			sgn = -1;
			c = snext();
		}
		long res = 0;
		do {
			if (c < '0' || c > '9')
				throw new InputMismatchException();
			res *= 10;
			res += c - '0';
			c = snext();
		} while (!isSpaceChar(c));
		return res * sgn;
	}

	public int[] nextIntArray(int n) {
		int a[] = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public String readString() {
		int c = snext();
		while (isSpaceChar(c))
			c = snext();
		StringBuilder res = new StringBuilder();
		do {
			res.appendCodePoint(c);
			c = snext();
		} while (!isSpaceChar(c));
		return res.toString();
	}

	public boolean isSpaceChar(int c) {
		if (filter != null)
			return filter.isSpaceChar(c);
		return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
	}

	public interface SpaceCharFilter {
		public boolean isSpaceChar(int ch);
	}
}
}
