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 __future__ import print_function # for PyPy2
from collections import Counter, OrderedDict
from itertools import permutations as perm
from fractions import Fraction
from collections import deque
from sys import stdin
from bisect import *
from heapq import *
from math import *

g   = lambda : stdin.readline().strip()
gl  = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
mod = int(1e9)+7
inf = float("inf")                      

n, x = gil()
a = gil()
b = Counter(a)
ans = 0
if x:
	for v in a:
		ans += b.get(x^v, 0)

	print(ans//2) 
else:
	for v in b.values():
		ans += (v*(v-1))//2
	print(ans)
