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>.
if (undefined == readline) {
	var fs = require('fs')
	var lines = fs.readFileSync('/dev/stdin', 'utf8').split("\n").reverse();
	var readline = function() {
		return lines.pop()
	}
}

if (undefined == print) {
	var print = console.log
}

var ints = function() {
	return readline().split(' ').map(function(n) {
		return parseInt(n)
	})
}

var numbers = ints(),
	x = numbers[1],
	counter = 0,
	temp = 0

numbers = ints().reduce(function(result, n) {
	if (undefined == result[n])
		result[n] = 0
	result[n] ++
	return result
}, {})

for (var n in numbers) {
	if (x == 0) {
		temp = numbers[n]
		if (temp > 1) {
			counter += (temp * (temp - 1)) / 2
		}
	} else {
		temp = numbers[n ^ x]
		if (undefined != temp)
			counter += temp * numbers[n]
		numbers[n] = undefined
	}
}

print(counter)