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;

typedef long long int ll;
typedef vector<int> vi;			
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef pair<int, int> 	pii;
typedef vector<bool> vb;
#define endl '\n'
#define all(x) x.begin(),x.end()
#define sz(x) (ll)x.size()
#define MOD 1000000007
#define F first
#define S second
#define pb push_back
#define pf push_front
#define rb pop_back
#define rf pop_front
#define in insert
#define mp make_pair
#ifdef LOCAL
	#include "../../trace.h"
#else
    #define trace(args...)
#endif

ll div_floor(const ll &a,const ll &b) { return a / b - (((a ^ b) < 0) and a % b);}
ll div_ceil(const ll &a,const ll &b) { return a / b + (((a ^ b) >= 0) and a % b);}

void solve(){
	ll n, k; cin>>n>>k;
	vl a(n); for(auto &e: a) cin>>e;

	ll lo = 0, hi = 1e9;
	while(lo+1<hi){
		ll mid = (lo + hi)/2;
		ll cntr = 0;
		for(auto &e: a) cntr += div_ceil(e, mid) - 1;
		if(cntr>k)
			lo = mid;
		else
			hi = mid;
	}
	cout<<hi<<endl;
}

int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	#ifdef LOCAL
		freopen(".inp.txt", "r", stdin);
    	freopen(".out.txt", "w", stdout);
    	freopen(".err.txt", "w", stderr);
	#endif
	int t=1;
	// cin>>t;
	while(t--)
		solve();
}