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.util.*;
import java.util.stream.Collectors;
import java.util.stream.Stream;

import static java.util.stream.Collectors.toList;

public class ProblemB {
    public static void main(final String[] args) {
        final Scanner sc = new Scanner(System.in);

        final int size = sc.nextInt();
        final int intComparator = sc.nextInt();

        final List<Integer> integersToCompare = new ArrayList<>(size);
        for (int i = 0; i < size; i++) {
            integersToCompare.add(i, sc.nextInt());
        }

        final long method = m2(intComparator, integersToCompare);

        System.out.println(method);
    }

    static long m2(final int intComparator, final List<Integer> integersToCompare) {
        final Map<Integer, Long> occurrences = integersToCompare
                .stream()
                .collect(Collectors.groupingBy(s -> s, Collectors.counting()));

        return integersToCompare
                .stream()
                .map(s -> new Val(s, s ^ intComparator))
                .filter(l -> occurrences.containsKey(l.corr))
                .map(val -> {
                    final int corr = val.corr;
                    if (val.val == corr) {
                        return occurrences.get(corr) - 1;
                    }
                    return occurrences.get(corr);
                })
                .reduce(0L, (a, b) -> a + b)
                / 2;
    }

    private static class Val {
        private final int val;
        private final int corr;


        private Val(final int val, final int corr) {
            this.val = val;
            this.corr = corr;
        }
    }
}
