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>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()
#define trav(a, x) for(auto& a : x)
typedef long long ll;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef long double ld;

int n,k;
vi a;

bool test(int m){
    int cuts = 0;
    rep(i,0,n){
        cuts += (a[i]-1) / m;
    }
    return cuts <= k;
}

int main(){
    cin.sync_with_stdio(0); cin.tie(0);
    cin.exceptions(cin.failbit);

    cin >> n >> k;
    a = vi(n);
    rep(i,0,n){
        cin >> a[i];
    }
    
    int lo = -1;
    int hi = 1e9+1;
    int mi;
    while(lo+1 < hi){
        mi = (lo+hi)/2;
        if(test(mi)){
            hi = mi;
        }else{
            lo = mi;
        }
    }
    cout << hi << endl;

    return 0;
}