Vasya has the sequence consisting of n integers. Vasya consider the pair of integers x and y k-interesting, if their binary representation differs from each other exactly in k bits. For example, if k = 2, the pair of integers x = 5 and y = 3 is k-interesting, because their binary representation x=101 and y=011 differs exactly in two bits.

Vasya wants to know how many pairs of indexes (i, j) are in his sequence so that i < j and the pair of integers a_i and a_j is k-interesting. Your task is to help Vasya and determine this number.

The first line contains two integers n and k (2 ≤ n ≤ 10⁵, 0 ≤ k ≤ 14) — the number of integers in Vasya's sequence and the number of bits in which integers in k-interesting pair should differ.

The second line contains the sequence a₁, a₂, ..., a_n (0 ≤ a_i ≤ 10⁴), which Vasya has.

Print the number of pairs (i, j) so that i < j and the pair of integers a_i and a_j is k-interesting.

In the first test there are 4 k-interesting pairs:

• (1, 3),
• (1, 4),
• (2, 3),
• (2, 4).

In the second test k = 0. Consequently, integers in any k-interesting pair should be equal to themselves. Thus, for the second test there are 6 k-interesting pairs:

• (1, 5),
• (1, 6),
• (2, 3),
• (2, 4),
• (3, 4),
• (5, 6).

用的新函数__builtin_popcount(i^j)计算i和j二进制不同的个数 用异或判断

#include<map>

#include<set>

#include<list>

#include<cmath>

#include<queue>

#include<stack>

#include<vector>

#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

#define pi acos(-1)

#define ll long long

#define mod 1000000007

using namespace std;

const int N=,maxn=+,inf=0x3f3f3f3f;

ll sum[maxn];

int main()

{

    ll n,k,ans=;

    memset(sum,,sizeof(sum));

    scanf("%I64d%I64d",&n,&k);

    for(int i=;i<n;i++)

    {

        ll a;

        scanf("%I64d",&a);

        sum[a]++;

    }

    if(!k)

    {

        for(int i=;i<=;i++)

            ans+=(sum[i]*(sum[i]-)/);

    }

    else

    {

        for(int i=;i<=;i++)

            for(int j=;j<i;j++)

               if(__builtin_popcount(i^j)==k)

                  ans+=(sum[i]*sum[j]);

    }

    printf("%I64d\n",ans);

    return ;

}

/*

4 1

0 3 2 1

4

*/