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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Main {

    static BufferedReader in;
    static PrintWriter out;
    static StringTokenizer tok;

    void solve() throws IOException {
        int n = ni(), x = ni();
        int[] a = nia(n);

        long[] cnt = new long[100001];
        for (int i = 0; i < n; i++) {
            cnt[a[i]]++;
        }

        long ans = 0;
        for (int i = 0; i <= 100000; i++) {
            int j = i ^ x;
            if (j > 100000) continue;
            if (i != j) {
                ans += cnt[i] * cnt[j];
            } else {
                ans += cnt[i] * (cnt[i] - 1);
            }
        }

        out.println(ans / 2);
    }

    String ns() throws IOException {
        while (!tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine(), " ");
        }
        return tok.nextToken();
    }

    int ni() throws IOException {
        return Integer.parseInt(ns());
    }

    long nl() throws IOException {
        return Long.parseLong(ns());
    }

    double nd() throws IOException {
        return Double.parseDouble(ns());
    }

    String[] nsa(int n) throws IOException {
        String[] res = new String[n];
        for (int i = 0; i < n; i++) {
            res[i] = ns();
        }
        return res;
    }

    int[] nia(int n) throws IOException {
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            res[i] = ni();
        }
        return res;
    }

    long[] nla(int n) throws IOException {
        long[] res = new long[n];
        for (int i = 0; i < n; i++) {
            res[i] = nl();
        }
        return res;
    }

    public static void main(String[] args) throws IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        tok = new StringTokenizer("");
        Main main = new Main();
        main.solve();
        out.close();
    }
}