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.math.BigInteger;
import java.util.*;

/**
 *
 * @author ramilagger
 */
public class Main {

    final static boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;
    char[] mask;

    void solve() {
        int n = nextInt();
        int x = nextInt();
        int[] a = nextArray(n);
        HashMap<Integer,Integer> map = new HashMap<>();
        for (int i = 0; i < n; i++) {
            if(map.containsKey(a[i]))
                map.put(a[i],map.get(a[i]) + 1);
            else
                map.put(a[i], 1);
        }
        long res = 0;
        for (int i = 0; i < n; i++) {
            if(map.containsKey(a[i] ^ x)) {
                if((a[i] ^ x) == a[i])
                res += map.get(a[i] ^ x) - 1;
                else res += map.get(a[i] ^ x);
            }
        }
        pw.println(res >>> 1);
    }


    public void run() throws IOException {
        start = System.currentTimeMillis();
        solve();
        if (!ONLINE_JUDGE)
            System.err.println(System.currentTimeMillis() - start + " ms");
        br.close();
        pw.close();
    }

    public Main() {
        try {
            br = (ONLINE_JUDGE) ? new BufferedReader(new InputStreamReader(System.in))
                    : new BufferedReader(new FileReader("in.txt"));
            pw = (ONLINE_JUDGE) ? new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)))
                    : new PrintWriter(new BufferedWriter(new FileWriter("out.txt")));
            this.run();
        } catch (Exception e) {
            e.printStackTrace();
            throw new RuntimeException();
        }
    }

    public static void main(String[] args) {
        new Main();
    }


    public boolean hasNext() {
        if (st != null && st.hasMoreTokens())
            return true;
        try {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
        } catch (Exception e) {
            return false;
        }
        return true;
    }

    public String next() {
        if (hasNext()) return st.nextToken();
        return null;
    }

    public int nextInt() {
        return Integer.parseInt(next());
    }

    public long nextLong() {
        return Long.parseLong(next());
    }

    public double nextDouble() {
        return Double.parseDouble(next());
    }

    public BigInteger nextBigInteger() {
        return new BigInteger(next());
    }

    public String nextLine() {
        StringBuilder sb;
        try {
            while (st == null || !st.hasMoreTokens()) return br.readLine();
            sb = new StringBuilder(st.nextToken());
            while (st.hasMoreTokens()) sb.append(" " + st.nextToken());
        } catch (IOException e) {
            throw new RuntimeException();
        }
        return sb.toString();
    }

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

    public long[] nextLArray(int n) {
        long[] temp = new long[n];
        for (int i = 0; i < n; i++)
            temp[i] = nextLong();
        return temp;
    }

    long start;
    final BufferedReader br;
    final PrintWriter pw;
    StringTokenizer st;

}