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.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.StringTokenizer;

public class Main {

    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Task solver = new Task();
        solver.solve(in, out);
        out.close();
    }
}

class Task {

    static int MAX = 1000001;

    public void solve(Scanner in, PrintWriter out) {
        int n = in.nextInt();
        int x = in.nextInt();
        int[] a = new int[n];
        int[] xor = new int[MAX];
        Arrays.fill(xor, 0);
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
        }
        long result = 0L;
        for (int i = 0; i < n; i++) {
            result += xor[a[i]];
            xor[a[i] ^ x]++;
        }
        System.out.println(result);
    }

}


class Utils {

    public static int binarySearch(int[] a, int key) {
        int s = 0;
        int f = a.length;
        while (f > s) {
            int mid = (s + f) / 2;
            if (a[mid] > key) {
                f = mid;
            } else if (a[mid] <= key) {
                s = mid + 1;
            }
        }
        return -1;
    }
}


class Scanner {
    BufferedReader in;
    StringTokenizer tok;

    public Scanner(InputStream in) {
        this.in = new BufferedReader(new InputStreamReader(in));
        tok = new StringTokenizer("");
    }

    private String tryReadNextLine() {
        try {
            return in.readLine();
        } catch (Exception e) {
            throw new InputMismatchException();
        }
    }

    public String nextToken() {
        while (!tok.hasMoreTokens()) {
            tok = new StringTokenizer(next());
        }
        return tok.nextToken();
    }

    public String next() {
        String newLine = tryReadNextLine();
        if (newLine == null)
            throw new InputMismatchException();
        return newLine;
    }

    public int nextInt() {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() {
        return Long.parseLong(nextToken());
    }
}