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.HashMap;
import java.util.Map;
import java.util.Scanner;
public class Main 
{
	public static void main(String[] args) 
	{
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int x = sc.nextInt();
		int[] a = new int[n];
		for (int i = 0; i < a.length; i++)
		{
			a[i] = sc.nextInt();
		}
		System.out.println(solve(a, x));
		sc.close();
	}
	static long solve(int[] a, int x) 
	{
		Map<Integer, Integer> valueToCount = new HashMap<>();
		for (int value : a) 
		{
			valueToCount.put(value, valueToCount.getOrDefault(value, 0) + 1);
		}
		long result = 0;
		for (int value : valueToCount.keySet())
		{
			int count = valueToCount.get(value);
			int other = value ^ x;
			if (other == value) 
			{
				result += count * (count - 1L);
			} 
			else 
			{
				result += (long) count * valueToCount.getOrDefault(other, 0);
			}
		}
		result /= 2;
		return result;
	}
}