## Algorithm

D2. Equalizing by Division (hard version)
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

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

You are given an array a consisting of n integers. In one move you can choose any ai�� and divide it by 22 rounding down (in other words, in one move you can set ai:=ai2��:=⌊��2⌋).

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

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

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

Input

The first line of the input contains two integers n and k (1kn21051≤�≤�≤2⋅105) — the number of elements in the array and the number of equal numbers required.

The second line of the input contains n integers a1,a2,,an�1,�2,…,�� (1ai21051≤��≤2⋅105), where ai�� is the i-th element of a.

Output

Print one integer — the minimum possible number of operations required to obtain at least k 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


## Code Examples

### #1 Code Example with C++ Programming

Code - C++ Programming

#include <bits/stdc++.h>

using namespace std;

int const N = 2e5 + 10;
int n, k, a[N], fr[N], frp[N];

int main() {
#ifndef ONLINE_JUDGE
freopen("in", "r", stdin);
#endif

scanf("%d %d", &n, &k);
for(int i = 0; i < n; ++i)
scanf("%d", a + i);
sort(a, a + n);

int res = 1e9;
for(int i = 0; i < n; ++i) {
int cur = a[i], cnt = 0;
while(cur >= 0) {
++fr[cur];
frp[cur] += cnt;
if(fr[cur] >= k)
res = min(res, frp[cur]);
if(cur == 0)
break;
cur /= 2;
++cnt;
}
}

printf("%d\n", res);

return 0;
}
Copy The Code &

Input

cmd
5 3
1 2 2 4 5

Output

cmd
1