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.util.*;

import static java.lang.Math.*;

public class Main2 {

    private FastScanner scanner = new FastScanner();

    public static void main(String[] args) {
        new Main2().solve();
    }

    private void solve() {

        int n = scanner.nextInt(), x = scanner.nextInt();

        int a[] = new int[n], d[] = new int[1000005];

        long cnt = 0;

        for (int i = 0; i < n; i++) {
            a[i] = scanner.nextInt();
            int b = x ^ a[i];
            cnt += d[b];
            d[a[i]] ++;
        }

        System.out.println(cnt);

    }

    boolean check(int cnt[][], int[] tcnt, int mid) {
        boolean ok = true;

        for (int j = 0; j < 27; j++) {
            if (cnt[mid][j] < tcnt[j]) {
                ok = false;
            }
        }
        return ok;
    }

    class Pair {
        int c, f;
    }

    class FastScanner {
        BufferedReader reader;
        StringTokenizer tokenizer;

        FastScanner() {
            reader = new BufferedReader(new InputStreamReader(System.in), 32768);
            tokenizer = null;
        }

        String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();

        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        Integer[] nextA(int n) {
            Integer a[] = new Integer[n];
            for (int i = 0; i < n; i++) {
                a[i] = scanner.nextInt();
            }
            return a;
        }

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

        public double nextDouble() {
            return Double.parseDouble(next());
        }

        public String nextLine() {
            try {
                return reader.readLine();
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }
    }
}