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>
using namespace std;

#define ll long long
// #define endl '\n'

void max_self(int &a, int b){ a = max(a, b);}
void min_self(int &a, int b){ a = min(a, b);}

int n, k; vector< int > v;

bool good(int x){
	int cnt = k;
	for(int i = 0; i<n; i++){
		cnt -= (v[i] - 1) / x;
	}
	return cnt >= 0;
}

void solve(){
	cin>>n>>k;
	v.resize(n);
	for(int &i : v) cin>>i;
	int l = -1; // bad
	int r = 1e9; // good

	while(r > l + 1){
		int mid = (l + r) / 2;
		if(good(mid)){
			r = mid;
		}else{
			l = mid;
		}
	}
	cout<<r<<endl;
}

int main(){
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int t = 1;// cin>>t;
	for(int i = 1; i<=t; i++){
		solve();
	}
	return 0;
}