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.StringTokenizer;

/*
br = new BufferedReader(new FileReader("input.txt"));
pw = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));
br = new BufferedReader(new InputStreamReader(System.in));
pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
 */

public class Main {
    private static BufferedReader br;
    private static StringTokenizer st;
    private static PrintWriter pw;

    public static void main(String[] args) throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));

        long[] occ = new long[100010];
        int n = readInt();
        int x = readInt();
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = readInt();
            occ[arr[i]]++;
        }

        long res = 0L;
        for(int num = 1; num <= 100000; num++) {
            int partner = num ^ x;
            if (num <= partner && partner <= 100000) {
                if (x == 0) {
                    res += (occ[num] * (occ[num] - 1L)) / 2L;
                } else {
                    res += (occ[num] * occ[partner]);
                }

            }
        }

        pw.println(res);
        pw.close();
    }


    private static void exitImmediately() {
        pw.close();
        System.exit(0);
    }

    private static long readLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    private static double readDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    private static int readInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    private static String nextToken() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
//            if (!br.ready()) {
//                exitImmediately();
//            }
            st = new StringTokenizer(br.readLine().trim());
        }
        return st.nextToken();
    }
}