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 random
#n,x=[random.randint(1,15),random.randint(1,10)]
#a=[random.randint(0,10) for i in range(n)]

n,x=map(int,raw_input().split())
a = map(int,raw_input().split())
def brute():
	ct=0
	for i in range(n):
		for j in range(i+1,n):
			if a[i]^a[j] == x:
				ct+=1


	return  ct

def opt():
	ct = 0
	checkArr = [0 for i in range(10**5)]
	length = max(a)+1
	for i in range(n):
		if checkArr[a[i]]==1:
			#print "checky",a[i],a[i]^x
			ct+=1
		newNum = a[i] ^ x
		while length <= newNum:
			checkArr.append(0)
			length+=1
		checkArr[newNum] = 1
	
	
	return ct

def opt1():
	ct = 0
	maxyNum = 0
	for i in range(n):
		if a[i]^x > maxyNum:
			maxyNum = a[i]^x
	checkArr = [0 for i in range(maxyNum+1)]
	length = max(a)+1
	for i in range(n):
		if a[i] <= maxyNum and checkArr[a[i]]>=1:
			#print "checky",a[i],a[i]^x
			ct+=checkArr[a[i]]
		newNum = a[i] ^ x
		'''
		while length <= newNum:
			checkArr.append(0)
			length+=1
		'''	
		checkArr[newNum] += 1
	
	
	return ct
'''
while True:
	if brute()!=opt1():
		print n,x
		print a
		print brute()
		print opt1()
		break
	else:
		print "correct"
'''
print opt1()
		
