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 Main {

    public static void main(String[] args) throws IOException {
        InputReader in = new InputReader();
        PrintWriter out = new PrintWriter(System.out);
        int t = 1;
        //        t = in.nextInt();
        Solver s = new Solver();
        while (t-- > 0) s.solve(in, out);
        out.close();
    }
}

class Solver {

    void solve(InputReader in, PrintWriter out) {
        int n = in.nextInt();
        int x = in.nextInt();
        HashMap<Integer, Integer> map = new HashMap<>();
        int[] a = new int[n];
        long ans = 0;
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
        }
        for (int i = n - 1; i >= 0; i--) {
            int e = a[i] ^ x;
            long size = 0;
            if (map.containsKey(e)) {
                size = map.get(e);
            }
            ans += (size);
            if (map.containsKey(a[i])) map.put(a[i], map.get(a[i]) + 1);
            else map.put(a[i], 1);
        }
        out.println(ans);
    }
}

class InputReader {
    BufferedReader br;
    StringTokenizer st;

    public InputReader() {
        br = new BufferedReader(new InputStreamReader(System.in));
        st = null;
    }

    public String next() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
        return st.nextToken();
    }

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

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

    public String nextLine() throws IOException {
        return br.readLine();
    }

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