codeforce vk cup2017

D. k-Interesting Pairs Of Integers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Input

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.

Output

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

Examples

Input

4 1
0 3 2 1

Output

4

Input

6 0
200 100 100 100 200 200

Output

6

Note

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
*/