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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.util.Map;
import java.util.HashMap;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        MyBufferedReader in = new MyBufferedReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskB solver = new TaskB();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskB {
        public void solve(int testNumber, MyBufferedReader in, PrintWriter out) {
            int[] data = in.getALineOfInts(2);
            int n = data[0];
            int x = data[1];
            int[] a = in.getALineOfInts(n);

            Map<Integer, Integer> withX = new HashMap<>();
            long res = 0;
            for (int i : a) {
                if (withX.containsKey(i)) {
                    res += withX.get(i);
                }

                int xor = i ^ x;
                if (withX.containsKey(xor)) {
                    withX.put(xor, withX.get(xor) + 1);
                } else {
                    withX.put(xor, 1);
                }
            }

            out.println(res);
        }

    }

    static class MyBufferedReader {
        BufferedReader in;

        public MyBufferedReader(InputStream s) {
            this.in = new BufferedReader(new InputStreamReader(s));
        }

        public int[] getALineOfInts(int numExpected) {
            int[] res = new int[numExpected];
            StringTokenizer st = null;
            try {
                st = new StringTokenizer(in.readLine());
            } catch (IOException e) {
                e.printStackTrace();
            }
            for (int i = 0; i < numExpected; i++)
                res[i] = Integer.parseInt(st.nextToken());
            return res;
        }

    }
}

