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;

#define int long long
#define endl '\n'

const int mxN = 2e5+5;
int a[mxN];
int n,k;
bool check(int m) {
    int ans = 0;
    for (int i = 0; i < n; i++) {
        int x = max(0LL, a[i]/m-1);
        if ((x+1)*m < a[i]) x++;
        ans += x;
    }
    if (ans <= k) return 1;
    return 0;
}
signed main(){
    ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    #ifdef LOCAL
    freopen("input.txt", "r" , stdin);
    freopen("output.txt", "w", stdout);
    #endif
    
    cin>>n;
    cin>>k;
    for (int i = 0; i < n; i++) {
        cin>>a[i];
    }
    int l = 1, r = 1e9;
    int ans = 1e9;
    while(l<=r) {
        int m = (l+r)/2;
        if (check(m)) {
            ans = m;
            r = m - 1;
        }
        else l = m + 1;
    }
    cout<<ans;
}