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)

#include<bits/stdc++.h>
#define int long long
#define FASTIO ios::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
using namespace std;
int n,k;
vector<int>v(n);
int valid(int x)
{
    int cnt=0;
    for(int i=0;i<n;i++)
    {
        if(v[i]<=x)
            continue;
        cnt+=(v[i])/x;
    }
    return cnt;
}
int32_t main()
{
    FASTIO;
    cin>>n>>k;
    int x;
    int mx=INT_MIN;
    for(int i=0;i<n;i++)
    {
        cin>>x;
        mx=max(mx,x);
        v.push_back(x);
    }
    int low=1;
    int high=mx;
    if(k==0)
    {
        cout<<mx;
        return 0;
    }
    int ans=-1;
    while(low<=high)
    {
        int mid=(low+high)/2;
        if(valid(mid)<=k)
        {
            ans=mid;
            high=mid-1;
        }
        else
            low=mid+1;
    }
    cout<<ans;
    return 0;
}