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;
#define int             ll
#define PI              3.1415926535897932384626433832795

#define pb              push_back
#define rep0            for(int i=0;i<n;i++)
#define rep1            for(int i=1;i<=n;i++)
#define inputar0(ar)    for(int i=0;i<n;i++)cin>>ar[i]
#define inputar1(ar)    for(int i=1;i<=n;i++)cin>>ar[i]

#define hell            freopen("input.txt","r", stdin);freopen("output.txt", "w", stdout);freopen("error.txt", "w",stderr);
#define fastasF_boi()   ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define endl            "\n"
//#define cerr            if(false)cerr

int mod=1000000007;
int maxn=200001;
int binpow(int a,int b,int mod){int res=1;while(b>0){if(b%2){res=(res*a)%mod;b--;}a=(a*a)%mod;b/=2;}return res%mod;}
int binpow(int a,int b){int res=1;while(b>0){if(b%2){res=(res*a);b--;}a=(a*a);b/=2;}return res;}
bool isPrime(int n){if(n==1)return false;for(int i=2;i*i<=n;i++){if(n%i==0)return false;}return true;}
int mulmod(int a,int b){return ((a%mod)*(b%mod))%mod;}
int addmod(int a,int b){return (((a%mod)+(b%mod))+mod)%mod;}

int ar[200005];

bool fun(int mid,int n,int k){
	int res=0;
	for(int i=0;i<n;i++){
        if(ar[i]>mid)
            res+=ar[i]/mid;
	}
	if(res<=k) return true;
	else return false;
}

void solve(){
    int n,k;cin>>n>>k;
	inputar0(ar);
	int l = 1,r=0,ans=-1;
	for(int i=0;i<n;i++){
        r=max(r,ar[i]);
	}
	while(l<=r){
		int mid = l+(r-l)/2;
		if(fun(mid,n,k)){
			r = mid-1;
			ans = mid;
		}
		else l = mid+1;
	}
	cout << ans;
}

signed main(){
//    hell;

    fastasF_boi();
    int t=1;
//    cin>>t;
    for(int tt=1;tt<=t;tt++){
        solve();
    }
}