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 Main {
    static FastScanner in;
    static int n, x;
    static int a[];
    static long ans = 0;
    static Map<Integer, Integer> map = new HashMap<Integer, Integer>();

    public static void main(String[] args) throws IOException {
        //      Scanner in = new Scanner(new File("input.txt"));
        //        Scanner in = new Scanner(System.in);
        //        in = new FastScanner("input.txt");
        in = new FastScanner(System.in);
        n = in.nextInt();
        x = in.nextInt();
        a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = in.nextInt();
            if (!map.containsKey(a[i]))
                map.put(a[i], 1);
            else
                map.put(a[i], map.get(a[i]) + 1);
        }
        for (int i = 0; i < n; i++) {
            int aj = x ^ a[i];
            if (map.containsKey(aj)) {
                if (aj == a[i])
                    ans += map.get(aj) - 1;
                else
                    ans += map.get(aj);
            }
        }
        System.out.println(ans / 2);
    }

}

class FastScanner {

    BufferedReader br;
    StringTokenizer tokenizer;

    FastScanner(String fileName) throws FileNotFoundException {
        this(new FileInputStream(new File(fileName)));
    }

    FastScanner(InputStream is) {
        br = new BufferedReader(new InputStreamReader(is));
    }

    String nextLine() throws IOException {
        tokenizer = null;
        return br.readLine();
    }

    String next() throws IOException {
        if (tokenizer == null || !tokenizer.hasMoreTokens()) {
            String line = br.readLine();
            if (line == null) {
                return null;
            }
            tokenizer = new StringTokenizer(line);
        }
        return tokenizer.nextToken();
    }

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

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

    char nextChar() throws IOException {
        return next().charAt(0);
    }
}