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>.
from sys import stdin,stdout,setrecursionlimit,maxint,exit
#setrecursionlimit(2*10**5)
def listInput():
 return map(long,stdin.readline().split())
def printBS(li):
 for i in xrange(len(li)-1):
  stdout.write("%d "%li[i])
 stdout.write("%d\n"%li[-1])
def sin():
 return stdin.readline().rstrip() 
n,x=listInput() 
li=listInput() 
dic={} 
for i in xrange(n): 
 if li[i]^x in dic: dic[li[i]^x].add(i) 
 else: dic[li[i]^x]=set([i])
ans=0
for i in xrange(n): 
 if li[i] in dic:
  ans+=len(dic[li[i]])
  if i in dic[li[i]]: ans-=1
print ans/2