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.Map;
import java.util.StringTokenizer;

public class B {

    final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;
    BufferedReader in;
    PrintWriter out;
    StringTokenizer tok = new StringTokenizer("");

    void solve() throws IOException {
        int n = readInt();
        int x = readInt();
        int[] a = readArr(n);
        int[] count = new int[100500];
        for(int i = 0; i < n; i++) {
            count[a[i]]++;
        }

        long res = 0;
        if(x == 0) {
            for(int i = 1; i < 100500; i++) {
                if(count[i] == 0) continue;
                res += count[i] * 1L * (count[i] - 1) / 2;
            }
            out.println(res);
        } else {
            for (int i = 1; i < 100500; i++) {
                int next = i ^ x;
                if (next < 1 || next > count.length - 1) {
                    continue;
                }
                res += count[i] * 1L * count[next];
            }
            res /= 2;
            out.println(res);
        }
    }
    void init() throws FileNotFoundException {
        if (ONLINE_JUDGE) {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
        } else {
            in = new BufferedReader(new FileReader("input.txt"));
            out = new PrintWriter("output.txt");
        }
    }

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

    int readInt() throws IOException {
        return Integer.parseInt(readString());
    }

    long readLong() throws IOException {
        return Long.parseLong(readString());
    }

    double readDouble() throws IOException {
        return Double.parseDouble(readString());
    }

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

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

    public static void main(String[] args) {
        new B().run();
    }

    public void run() {
        try {
            long t1 = System.currentTimeMillis();
            init();
            solve();
            out.close();
            long t2 = System.currentTimeMillis();
            System.err.println("Time = " + (t2 - t1));
        } catch (Exception e) {
            e.printStackTrace(System.err);
            System.exit(-1);
        }
    }
}