Problem:
Score : 500 points

Problem Statement

We have N logs of lengths A_1,A_2,\cdots A_N.

We can cut these logs at most K times in total. When a log of length L is cut at a point whose distance from an end of the log is t (0<t<L), it becomes two logs of lengths t and L-t.

Find the shortest possible length of the longest log after at most K cuts, and print it after rounding up to an integer.

Constraints

1 \leq N \leq 2 \times 10^5

0 \leq K \leq 10^9

1 \leq A_i \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 \cdots A_N

Output

Print an integer representing the answer.

Sample Input 1

2 3
7 9

Sample Output 1

4

First, we will cut the log of length 7 at a point whose distance from an end of the log is 3.5, resulting in two logs of length 3.5 each.

Next, we will cut the log of length 9 at a point whose distance from an end of the log is 3, resulting in two logs of length 3 and 6.

Lastly, we will cut the log of length 6 at a point whose distance from an end of the log is 3.3, resulting in two logs of length 3.3 and 2.7.

In this case, the longest length of a log will be 3.5, which is the shortest possible result. After rounding up to an integer, the output should be 4.

Sample Input 2

3 0
3 4 5

Sample Output 2

5

Sample Input 3

10 10
158260522 877914575 602436426 24979445 861648772 623690081 433933447 476190629 262703497 211047202

Sample Output 3

292638192

Solution in C++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
#define ll long long int
using namespace std;
const int mod=1e9+7;
// Not for Rating """"""""""I'm in Love with Experience :)""""""""""""""""""
bool logs(ll *A,ll N,ll K,ll mid)
{
    ll a=0;
    for(ll i=0;i<N;++i)
    {
        if(A[i]<=mid) continue;
        a+= (A[i]+mid - 1)/mid    - 1;
    }
    return a<=K;
}
void solve()
{
    ll N;cin>>N;ll K;cin>>K;
    ll A[N],l=0; 
    for(ll i=0;i<N;++i)
    {
        cin>>A[i];
        //h+=A[i];
    }
    ll mx = *max_element(A,A+N);
    
    ll ans=mx,h=mx;
    while(l<=h)
    {
        ll mid = l + (h-l)/2;
        if(logs(A,N,K,mid))
        {
            ans=min(ans,mid);
            h=mid-1;
        }
        else l=mid+1;
    }
    cout<<ans;
}
int main(void)
{
    ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
    int T=1;while(T--){solve();}return 0;
}/*Solved By:- RITIK AGARWAL*/