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 CF_A implements Runnable {

    private static final boolean ONLINE_JUDGE = System.getProperty("ONLINE_JUDGE") != null;

    private BufferedReader in;
    private PrintWriter out;
    private StringTokenizer tok = new StringTokenizer("");

    private void init() throws FileNotFoundException {
        Locale.setDefault(Locale.US);
        String fileName = "";
        if (ONLINE_JUDGE && fileName.isEmpty()) {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
        } else {
            if (fileName.isEmpty()) {
                in = new BufferedReader(new FileReader("input.txt"));
                out = new PrintWriter("output.txt");
            } else {
                in = new BufferedReader(new FileReader(fileName + ".in"));
                out = new PrintWriter(fileName + ".out");
            }
        }
    }

    String readString() {
        while (!tok.hasMoreTokens()) {
            try {
                tok = new StringTokenizer(in.readLine());
            } catch (Exception e) {
                return null;
            }
        }
        return tok.nextToken();
    }

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

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

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

    int[] readIntArray(int size) {
        int[] a = new int[size];
        for (int i = 0; i < size; i++) {
            a[i] = readInt();
        }
        return a;
    }

    public static void main(String[] args) {
        //new Thread(null, new _Solution(), "", 128 * (1L << 20)).start();
        new CF_A().run();
    }

    long timeBegin, timeEnd;

    void time() {
        timeEnd = System.currentTimeMillis();
        System.err.println("Time = " + (timeEnd - timeBegin));
    }

    @Override
    public void run() {
        try {
            timeBegin = System.currentTimeMillis();
            init();
            solve();
            out.close();
            time();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(-1);
        }
    }

    int[] tree;
    int ans = 0;


    void solve() throws IOException {
        int n = readInt();


        int m = readInt();
        HashMap<Integer, Integer> hm = new HashMap<>();
        for (int i = 0; i < 100000; i++) {
            hm.put(i + 1, 0);
        }
        for (int i = 0; i < n; i++) {
            int k = readInt();
            hm.put(k, hm.get(k) + 1);
        }
        long sum = 0;
        long buf = 0;
        for (int i = 1; i <= hm.size(); i++) {
            int p = i ^ m;
            if(hm.containsKey(i ^ m)) {
                if(p != i) {
                    long k = (long) hm.get(i) * (long) hm.get(p);
                    sum += k;

                }else{
                    buf+=((long)hm.get(i))*(hm.get(i)-1)/2;
                }
            }
        }
        out.println(sum / 2 + buf);


    }


}