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≤2⋅1051≤k≤n≤2⋅105) — 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

The first line of the input contains two integers nn and kk (1≤k≤n≤2⋅1051≤k≤n≤2⋅105) — 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.

Exanple

Input

5 3
1 2 2 4 5

Output

1

Input

5 3
1 2 3 4 5

Output

2

Input

5 3
1 2 3 3 3

Output

0

简单用中文描述一下就是使一组数通过除2，来达到有k个相同的值的结果，然后问你最少要通过几步   思路就是创建一个优先队列，里面存放的是可以变成他的数，变成他所需要的步数，然后奖前k项相加，输出最小值就可以了

#include<iostream>
#include<algorithm>
#include<queue>
#include<cstdio> 
#define INF 0x3f3f3f3f
using namespace std;

priority_queue<int,vector<int>,greater<int> > q[(int)3e5];

int main(){
    int x,y,a;
    scanf("%d%d",&x,&y);
    for(int i=0;i<x;i++){
        scanf("%d",&a);
        q[a].push(0);
        int cnt=1;
        while(a){    
            a/=2;
            q[a].push(cnt);
            cnt++;
        }
    }
    
    int asn=INF;
    for(int i=0;i < 3e5;i++){
        if(q[i].size()<y)
            continue;
        else{
            int sum=0;
            for(int j=0;j<y;j++){
                    sum=sum+q[i].top();
                    q[i].pop();
            }
            asn=min(asn, sum);
    }
}

    printf("%d\n",asn);
    return 0;
}