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 endl '\n'
#define ll long long
#define all(a) a.begin(), a.end()
#define sz(a) a.size()
using namespace std;

int n, k, a[200001], ans;

bool ok(int x){
    int tk = k;
    for(int i=0; i<n; i++){
        tk -= (a[i]-1)/x;
        if(tk<0) return false;
    }
    return true;
}

int main()
{
    ios_base::sync_with_stdio(0); cin.tie(0);
    #ifndef ONLINE_JUDGE
    //freopen("A.txt", "r", stdin);
    #endif
    cin >> n >> k;
    for(int i=0; i<n; i++){
        cin >> a[i];
        ans = max(ans, a[i]);
    }
    int lb = 0, rb = ans;
    while(lb<rb){
        int mid = (lb+rb)/2;
        if(ok(mid)){
            ans = mid;
            rb = mid;
        }
        else lb = mid+1;
    }
    cout << ans << endl;
}