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.math.*;  //BigInteger, BigDecimal
import java.util.*;  //StringTokenizer, ArrayList


public class R383_Div2_B
{	
	FastReader in;
	PrintWriter out;
	
	public static void main(String[] args)  {
		new R383_Div2_B().run();
	}
	
	void run()
	{		
		in = new FastReader(System.in);
		out = new PrintWriter(System.out);
		solve();
		out.close();
	}
	
	void solve()
	{
		int n = in.nextInt();
		int x = in.nextInt();
		
	    int[] d = new int[20];
	    int it2 = x;
	    for (int i = 0; i < 20; i++)
	    {
	        d[i] = it2 % 2;
	        it2 /= 2;
	    }
		
	    int[] num = new int[100_001];
		int[] a = new int[n];
		for (int i = 0; i < n; i++) 
		{
			a[i] = in.nextInt();
			num[a[i]]++;
		}
		
		long tot = 0;
		for (int i = 1; i < 100_001; i++)
			if (num[i] > 0)
			{
				int xor = i ^ x;
				if (xor <= 100_000 && xor <= i && num[xor] > 0)
				{
					if (i == xor) 
						tot += (1L * num[i] * (num[i] - 1)) / 2L;
					else
						tot += 1L * num[i] * num[xor];						
				}
			}
		
		out.println(tot);		
	}

	//-----------------------------------------------------
	void runWithFiles() {
		in = new FastReader(new File("input.txt"));
		try {
			out = new PrintWriter(new File("output.txt"));
		} 
		catch (FileNotFoundException e) {
			e.printStackTrace();
		}
		solve();
		out.close();
	}
	
	class FastReader
	{
	    BufferedReader br;
	    StringTokenizer tokenizer;
	    
	    public FastReader(InputStream stream)
	    {
	        br = new BufferedReader(new InputStreamReader(stream));
	        tokenizer = null;
	    }
		public FastReader(File f) {
			try {
				br = new BufferedReader(new FileReader(f));
				tokenizer = null;
			}
			catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}
	    
	    private String next() {
	        while (tokenizer == null || !tokenizer.hasMoreTokens())
	            try {
	            	tokenizer = new StringTokenizer(br.readLine());
	            }
	            catch (IOException e) {
	                throw new RuntimeException(e);
	            }
	        return tokenizer.nextToken();
	    }
		public String nextLine() {
			try	{
				return br.readLine();
			}
			catch(Exception e) {
				throw(new RuntimeException());
			}
		}

	    int nextInt() {
	        return Integer.parseInt(next());
	    }
	    long nextLong() {
	        return Long.parseLong(next());
	    }
	    double nextDouble() {
	        return Double.parseDouble(next());
	    }	    
	    BigInteger nextBigInteger() {
	        return new BigInteger(next());
	    }
	    BigDecimal nextBigDecimal() {
	        return new BigDecimal(next());
	    }
	}
}
