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)

//NEVER GIVE UP I AM GOING TO IMPROVE ONE DAY.
//PRACTICE,PRACTICE,PRACTICE!!!
#include<bits/stdc++.h>
#define int long long int
// #define ull unsigned long long int
#define pb push_back
#define mp make_pair
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 1e18
#define endl "\n"
#define all(v) v.begin(),v.end()
#define fast ios_base::sync_with_stdio(false); cin.tie(0);cout.tie(0); 
#define dbg(x)  cerr << #x << " = " << x << endl;
const int M=100005;
const int mod=1000000007;
const double pi = acos(-1.0);
using namespace std;
bool possible(int mid,int k,int a[],int n){
    int cut=0;
    for(int i=0;i<n;i++){
        cut+=((a[i]-1)/mid);
    }
    if(cut<=k){
        return true;
    }
    return false;
}
int32_t main(){
 
    // #ifndef ONLINE_JUDGE 
    //     freopen("in.txt", "r", stdin); 
    //     freopen("out.txt", "w", stdout); 
    // #endif 
 
    fast;
    int t;
    t=1;
    // cin>>t;
    // int test=t;
    while(t--){
        int n,k;
        cin>>n>>k;
        int a[n],maxi=-1;
        for(int i=0;i<n;i++){
            cin>>a[i];
            maxi=max(maxi,a[i]);
        }
        int low=0,high=maxi,mid,ans=maxi;
        while(high>=low){
            mid=low+((high-low)/2);
            if(possible(mid,k,a,n)){
                ans=min(ans,mid);
                high=mid-1;
            }
            else{
                low=mid+1;
            }
        }
        cout<<ans<<endl;
    }
}