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>
#define ll long long
#define pb push_back
#define mp make_pair
#define F first
#define S second
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pcc pair<char,char>
#define vi vector <int>
#define vl vector <ll>
#define sd(x) scanf("%d",&x)
#define slld(x) scanf("%lld",&x)
#define pd(x) printf("%d",x)
#define plld(x) printf("%lld",x)
#define pds(x) printf("%d ",x)
#define pllds(x) printf("%lld ",x)
#define pdn(x) printf("%d\n",x)
#define plldn(x) printf("%lld\n",x)
using namespace std;
const int INF = 2e9;
const ll INFLL = 4e18;
ll powmod(ll base,ll exponent,ll mod){ // with mod < 1e9
	ll ans=1;
	while(exponent){
		if(exponent&1)ans=(ans*base)%mod;
		base=(base*base)%mod;
		exponent/=2;
	}
	return ans;
}
ll gcd(ll a, ll b){
	if(b==0) return a;
	else return gcd(b,a%b);
}
const int upperlimit = 1e6+100;
const int mod = 1e9+7;
int a[upperlimit];
int n;
int cuts(int l){
	int ans=0;
	for(int i = 1; i <= n; i++) ans+=(a[i]/l);
	return ans;
}
int main() {
	int k,l=0,r=INF;
	sd(n);sd(k);
	for(int i = 1; i <= n; i++){sd(a[i]);a[i]--;}
	while(r-l>1){
		int mid=(l+r)>>1;
		if(cuts(mid)<=k) r=mid;
		else l=mid;
	}
	pd(r);
	return 0;
}