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.lang.reflect.Array;
import java.util.*;

public class Main {


    public void solve() throws IOException {
        int n = nextInt (), x = nextInt ();
        TreeMap<Integer, Integer> a = new TreeMap<> ();
        for (int i = 0; i < n; i++) {
            int y = nextInt ();
            if (a.containsKey (y)) {
                a.put (y, a.get (y) + 1);
            } else a.put (y, 1);
        }
        long res = 0;
        for (Integer i : a.keySet ()) {
            if (a.containsKey (i ^ x)) {
                res += (long) a.get (i ^ x) * a.get (i);
                if(x == 0) res -= a.get (i);
            }
        }
        res /= 2;
        out.print (res);
    }


    BufferedReader br;
    StringTokenizer sc;
    PrintWriter out;

    public String nextToken() throws IOException {
        while (sc == null || !sc.hasMoreTokens ()) {
            try {
                sc = new StringTokenizer (br.readLine ());
            } catch (Exception e) {
                return null;
            }
        }
        return sc.nextToken ();
    }

    public int nextInt() throws IOException {
        return Integer.parseInt (nextToken ());
    }

    public boolean hasNext() {
        while (sc == null || !sc.hasMoreTokens ()) {
            try {
                String s = br.readLine ();
                if (s == null) {
                    return false;
                }
                sc = new StringTokenizer (s);
            } catch (IOException e) {
                throw new RuntimeException (e);
            }
        }
        return sc.hasMoreTokens ();
    }

    public double nextDouble() throws IOException {
        return Double.parseDouble (nextToken ());
    }

    public long nextLong() throws IOException {
        return Long.parseLong (nextToken ());
    }

    public static void main(String[] args) throws IOException {
        Locale.setDefault (Locale.US);
        new Main ().run ();
    }

    public void run() {
        try {
            out = new PrintWriter (System.out);
            br = new BufferedReader (new InputStreamReader (System.in));
//            br = new BufferedReader (new FileReader ("input.txt"));
//            out = new PrintWriter (new File ("output.txt"));
            solve ();
            out.close ();
        } catch (IOException e) {
            e.printStackTrace ();
            System.exit (1);
        }
    }
}