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>.
n, x=[int(i) for i in input().split()]
a=[int(i) for i in input().split()]

ax=[i^x for i in a]
a.sort()
ax.sort()
pos=0
posx=0
c=1
cx=1
total=0
while pos<n and posx<n:
	if a[pos]==ax[posx]:
		if pos<n-1 and a[pos+1]==a[pos]:
			pos+=1
			c+=1
		elif posx<n-1 and ax[posx+1]==ax[posx]:
			posx+=1
			cx+=1
		else:
			#print(c, cx, pos, posx)
			total+=c*cx
			c=1
			cx=1
			pos+=1
			posx+=1
	elif a[pos]>ax[posx]:
		posx+=1
	else:
		pos+=1
#print(total, a, ax)
if x>0:
	print(total//2)
else:
	print((total-n)//2)


