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: Accepted)

/*
    Author: Alam Khan
    AUST CSE 40th Batch

*/

#include<bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/detail/standard_policies.hpp>
using namespace __gnu_pbds;

typedef long long ll;
typedef tree<ll,null_type,less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update>ordered_set;
#define pb     push_back
#define inf     2e18
#define low     -2e18
#define PI        acos(-1.0)
#define endl  "\n"
#define FAST  ios_base::sync_with_stdio(false);cin.tie(NULL);
#define fr freopen("input.txt","r",stdin)
#define fw freopen("output.txt","w",stdout)
const int sz =2e5+9;
ll a[sz];
int main()
{
    ll i,n,t,k,j,x=0,y=0,m;
    cin>>n>>k;
    for(i=0;i<n;i++)
    {
        cin>>a[i];
    }
    ll lo,mid,hi;
    ll ans = 0;
    lo = 1;
    hi = 1e9;
    while(lo<=hi)
    {
        mid = (lo+hi)/2;
        x = 0;
        for(i=0;i<n;i++)
        {
            if(a[i]%mid!=0)
                y = a[i]/mid+1;
            else y = a[i]/mid;
            x+= y-1;
        }
        if(x<=k)
        {
            ans = mid;
            hi = mid-1;
        }
        else lo  =mid+1;
    }
    cout<<ans<<endl;
    return 0;
}