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>
#define mod 1000000007
#define ll long long int
#define pb push_back
#define mk make_pair
#define pii pair<int,int>
#define pll pair<ll,ll>
#define fr(j,i,n) for(int i=j;i<n;++i)
#define frr(j,i,n) for(int i=j;i<=n;++i)
using namespace std;
void solve()
{
	ll n,k;
	cin>>n>>k;
	ll a[n+1];
	frr(1,i,n)	cin>>a[i];
	ll left=0,right=1e9;
	ll ans=1e10;
	while(left<=right)
	{
		ll moves=k;
		ll mid=(left+right)/2ll;
		frr(1,i,n)
		{
			if(a[i]%mid==0)
				moves-=((a[i]/mid)-1ll);
			else
				moves-=(a[i]/mid);
		}
		if(moves>=0)
			ans=min(mid,ans),right=mid-1ll;
		else
			left=mid+1ll;
	}
	cout<<ans<<"\n";
}
int main()
{
	// #ifndef ONLINE_JUDGE
	// 	freopen("input.txt","r",stdin);
	// 	freopen("output.txt","w",stdout);
	// #endif
	ios_base::sync_with_stdio(false);
    cin.tie(NULL);
	int t=1;
	// t=1;
	// cin>>t;
	while(t--)
	solve();
	return 0;
}