codeforces Equalizing by Division (easy version)

output

standard output

The only difference between easy and hard versions is the number of elements in the array.

You are given an array aa consisting of nn integers. In one move you can choose any aiai and divide it by 22 rounding down (in other words, in one move you can set ai:=⌊ai2⌋ai:=⌊ai2⌋).

You can perform such an operation any (possibly, zero) number of times with any aiai.

Your task is to calculate the minimum possible number of operations required to obtain at least kk equal numbers in the array.

Don't forget that it is possible to have ai=0ai=0 after some operations, thus the answer always exists.

Input

The first line of the input contains two integers nn and kk (1≤k≤n≤501≤k≤n≤50) — the number of elements in the array and the number of equal numbers required.

The second line of the input contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤2⋅1051≤ai≤2⋅105), where aiai is the ii-th element of aa.

Output

Print one integer — the minimum possible number of operations required to obtain at least kk equal numbers in the array.

Examples

input

Copy

5 3
1 2 2 4 5

output

Copy

1

input

Copy

5 3
1 2 3 4 5

output

Copy

2

input

Copy

5 3
1 2 3 3 3

output

Copy

0
题解：开一个数组cnt，保存某一个数出现的次数。然后循环除2，同时再开一个数组num,用来保存当初的数字，需要除多少次2才能变成当前的值。同时判断当前数字出现的次数，

#include<bits/stdc++.h>
using namespace std;
const int N=2E5+;
const int inf=1e9+;
int arr[N];
int num[N];
int cnt[N];
int main(){
    int n,k;
    cin>>n>>k;
    for(int i=;i<=n;i++) scanf("%d",&arr[i]);
    sort(arr+,arr++n);
    int ans=inf;
    for(int i=;i<=n;i++){
        int temp=;
        int x=arr[i];
        while(x){
            cnt[x]++;//判断arr[i]出现的次数;
            num[x]+=temp;//判断arr[i]编程现在的arr[i]需要操作几次;
            if(cnt[x]==k) {
                ans=min(ans,num[x]);
            }
            temp++;
            x/=;
        }
    }
    printf("%d",ans);
    return ;
}