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>
using namespace std;
#define int long long
#define FASTIO ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define pii pair<int, int>
#define ff first
#define ss second
#define PI acos(-1)
#define ld long double
//mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
const int mod = 1e9+7, N = 2e5+5;

int a[N], n, k;
bool check(int m){
	int c = 0;
	for(int i=0;i<n;i++){
		if(a[i]<=m)continue;
		c += (floor)(a[i]-1)/m;
	}
	return (c <= k);
}
void solve(int test_case){
	int i, j;
	int mx = 0;
	cin >> n >> k;
	for(i=0;i<n;i++){
		cin >> a[i];
		mx = max(a[i], mx);
	}
	if(k == 0){
		cout << mx;
		return;
	}
	int l = 0, r = mx;
	while(l+1 < r){
		int mid = (l+r)/2;
		if(check(mid)){
			r = mid;
		}else {
			l = mid;
		}
	}
	cout << r;
	return;
}

signed main(){
	FASTIO;
#define MULTITEST 0
#if MULTITEST
	int ___T;
	cin >> ___T;
	for(int T_CASE = 1; T_CASE <= ___T; T_CASE++)
		solve(T_CASE);
#else
	solve(1);
#endif
	return 0;
}