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)

/**
 ____ ____ ____ ____ ____
||a |||t |||o |||d |||o ||
||__|||__|||__|||__|||__||
|/__\|/__\|/__\|/__\|/__\|

**/

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int N_MAX = 200002;

int n;
ll k;

ll a[N_MAX];

bool f (int mx)
{
    ll cnt = 0;
    for(int i = 1; i <= n; i++)
    {
        cnt += (a[i] + mx - 1) / mx - 1;
        if(cnt > k)
            return false;
    }
    return true;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> k;
    for(int i = 1; i <= n; i++)
        cin >> a[i];
    ll l = 0, r = 1000000000;
    while(l < r)
    {
        ll mid = ((l + r) >> 1);
        if(f(mid))
            r = mid;
        else
            l = mid + 1;
    }
    cout << l << "\n";
    return 0;
}