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.HashMap;
import java.util.StringTokenizer;

public class pr742B {
    public static void main(String[] args) throws IOException {
        FastScanner in = new FastScanner(System.in);
        int n = in.nextInt();
        int x = in.nextInt();
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
        }

        PrintWriter out = new PrintWriter(System.out);
        out.println(solve(n, x, a));
        out.flush();
        out.close();
    }

    private static long solve(int n, int x, int[] a) {
        long ans = 0;
        HashMap<Integer, Integer> hm = new HashMap<>();
        for(int i = 0; i < n; i++) {
            int y = x ^ a[i];
            if(hm.containsKey(y)) ans += hm.get(y);
            if(hm.containsKey(a[i])) hm.put(a[i], hm.get(a[i])+1);
            else hm.put(a[i], 1);
        }
        return ans;
    }

    static class FastScanner {
        StringTokenizer st;
        BufferedReader br;

        FastScanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        boolean hasNext() throws IOException {
            return br.ready() || (st != null && st.hasMoreTokens());
        }

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

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

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

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

        boolean hasNextLine() throws IOException {
            return br.ready();
        }

    }
}
