codeforcesB - Arpa’s obvious problem and Mehrdad’s terrible solution 【思维】

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


, where


is bitwise
xor operation (see notes for explanation).



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.

Example

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.

so the answer is
1.

In the second sample the only two pairs are i = 3,
j = 4 (since

) and
i = 1, j = 5 (since


).

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.
题意：Given an array and a number
x, count the number of pairs of indices i, j (1 ≤ i < j ≤ n) such that


, where


is bitwise
xor operation

总共100000个数啊，咋找啊，想到拆位，每个数拆成二进制最多是17位，但是啊，这么做复杂度依旧降不下来啊。

需要注意到，位异或有一个特殊的性质 a^b=c那么a^c=b  b^c=a所以说我想找有多少对数可以位异或等于x，那么我就求num[i]^x是否在num[1-n]中出现过就好啦~~出现过几个最后加几个结果

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int exi[200009];
int num;
int main()
{
int n,x;
while(cin>>n>>x)
{
memset(exi,0,sizeof(exi));
long long tot=0;
for(int i=1;i<=n;i++)
{
scanf("%d",&num);
if(exi[num^x]) tot+=exi[num^x];
exi[num]=exi[num]+1LL;
}
cout<<tot<<endl;
}
return 0;
}