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, ..., a_n (1 ≤ a_i ≤ 10⁵) — the elements of the array.

Output

Print a single integer: the answer to the problem.

Example

Input

2 31 2

Output

1

Input

6 15 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.

这道题一开始没看懂啥意思，后来才明白原来是异或^，a^b=c，那么a^c=b；抓住这一点就好做了，所有的ai去异或x得到cnt，然后记录cnt，看看数组里面有几个cnt就好了，可以开个map，注意结果可能超过long 要用long long

代码：

#include <iostream>
#include <cstdio>
#include <map>
using namespace std;
int n,x,a[100000];
map<int,int> mark;
void input()
{
    cin>>n>>x;
    for(int i=0;i<n;i++)
    {
        cin>>a[i];
        mark[a[i]]++;
    }
}
long long check()
{
    long long cnt,ans=0;
    for(int i=0;i<n;i++)
    {
        cnt=a[i]^x;
        ///x如果是0 那么 任意一个数q q^0=q;
        if(a[i]==cnt)ans+=mark[cnt]-1;
        else ans+=mark[cnt];
    }
    return ans/2;
}
int main()
{
    input();
    cout<<check();
}