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>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <functional>
using namespace std;
using namespace __gnu_pbds;

#define ll long long
#define mk make_pair

template <class key, class cmp = std::less<key>>
using ordered_set = tree<key, null_type, cmp, rb_tree_tag, tree_order_statistics_node_update>;
const ll mod = 1e9 + 7;

/*
    if you have any question about any part of my code,
    please don't shy to ask anytime :)
*/

ll n,k;
ll x[200005];

bool get(ll mid)
{
    ll cnt=0;
    for(int i=0;i<n;i++)
    {
        cnt+=(x[i]-1)/mid;
    }

    return cnt<=k;
}

void solve()
{
    cin>>n>>k;
    for(int i=0;i<n;i++)
    {
        cin>>x[i];
    }

    ll st=1, en=1e9;
    ll ans=0;
    while(st<=en)
    {
        ll mid = (st+en)>>1;
        if(get(mid))
        {
            ans = mid;
            en = mid-1;
        }
        else
        {
            st = mid+1;
        }
    }

    cout<<ans<<endl;
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    // ll t;
    // cin >> t;
    // while (t--)
    // {
    //     solve();
    // }
    solve();
    return 0;
}