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 {
    public static void main(String[] args) throws FileNotFoundException, IOException {
        InputStream inputStream = System.in;
//        InputStream inputStream = new FileInputStream("dijkstra.in");
        OutputStream outputStream = System.out;
//        OutputStream outputStream = new FileOutputStream("dijkstra.out");
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Answer solver = new Answer();
        solver.solve(1, in, out);
        out.close();
    }

    static class Answer {
        int find(List<Integer> a, int key, int l, int r) {
            if (l > r) {
                return -1;
            }
            int m = (l + r) >> 1;
            if (a.get(m) < key) {
                return find(a, key, m + 1, r);
            }

            if (a.get(m) > key) {
                int x = find(a, key, l, m - 1);
                if (x != -1) {
                    return x;
                }
                return m;
            }
            return m;
        }

        public void solve(int testNumber, InputReader in, PrintWriter out) throws IOException {
            int n = in.nextInt();
            int x = in.nextInt();

            List<Integer> l = new ArrayList<>();
            List<Integer>[] count = new List[100001];
            for (int i = 0; i < count.length; i++) {
                count[i] = new ArrayList<>();
            }

            for (int i = 0; i < n; i++) {
                int a = in.nextInt();
                l.add(a);
                count[a].add(i);
            }

            long ans = 0;
            for (int i = 0; i < n; i++) {
                int y = x ^ l.get(i);
                if (y > 100000) {
                    continue;
                }
                if (count[y].size() > 0) {
                    int xx = find(count[y], i + 1, 0, count[y].size() - 1);
                    if (xx != -1) {
                        ans += count[y].size() - xx;
                    }
                }
            }

            out.print(ans);
        }
    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

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

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

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