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 $\text{[math]}$ , where $\text{[math]}$ 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, ..., a_n (1 ≤ a_i ≤ 10⁵) — the elements of the array.

Output

Print a single integer: the answer to the problem.

Example

Input

2 3
1 2

Output

Input

6 1
5 1 2 3 4 1

Output

Note

In the first sample there is only one pair of i = 1 and j = 2. $\text{[math]}$ so the answer is 1.

In the second sample the only two pairs are i = 3, j = 4 (since $\text{[math]}$ ) and i = 1, j = 5 (since $\text{[math]}$ ).

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 $\text{[math]}$ , where $\text{[math]}$ 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;
}