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: Wrong Answer)

#include <bits/stdc++.h>
#include <iostream>
using namespace std;
#define ll long long
#define ld long double
#define mp make_pair
#define pb push_back
#define fo(i,n) for(ll i=0;i<n;i++)
#define fo1(i,n) for(ll i=1;i<=n;i++)
#define loop(i,a,b)for(ll i=a;i<=b;i++)
#define loopr(i,a,b)for(ll i=b;i>=a;i--)
#define all(x) x.begin(), x.end()
#define sz(x) (ll)(x).size()
#define vll vector<ll> 
#define vvl vector<vll>
#define vpll vector<pll>
#define pll pair<ll,ll>
#define F first
#define S second
#define MOD 1000000007
ll max(ll a,ll b){if (a>b) return a; else return b;}
ll gcd(ll a, ll b){if(b==0)return a;return gcd(b, a%b);}
ll lcm(ll a, ll b){return a*b/gcd(a, b);}
ll fexp(ll a, ll b){ll ans = 1;while(b){if(b&1) ans = ans*a%MOD; b/=2;a=a*a%MOD;}return ans;}
ll inverse(ll a, ll p){return fexp(a, p-2);}
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	ll T=1;
	//cin >> T;
	ll t=0;
	while(t++<T){
		ll n,k;
		cin>>n>>k;
		vll a(n);
		fo(i,n)cin>>a[i];
		ll lo=1,hi=*max_element(all(a)),ans;
		ll m;
		while(lo<=hi){
			m=(lo+hi)/2;
			ll req=0;
			fo(i,n){
				req+=2*a[i]/(2*m+1);
			}
			if(req<=k){ans=m;hi=m-1;}
			else lo=m+1;
		}
		cout<<ans<<endl;
	}
return 0;
}