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 lli long long
#define ld long double
#define mod 1000000007 
#define set(n) cout << fixed << setprecision(n)    
#define fi first
#define se second
#define mp make_pair
#define pll pair<lli, lli>
#define endl '\n'

const double PI = acos(-1);
const double epsilon = 1e-7;  // floating point precision

bool check(int minSize, vector<int>& v, int k) {
    int kk = k;
    for(int i = 0; i < v.size(); i++) {
        kk -= (v[i] / minSize  - 1 + min(1, v[i] % minSize));
        if(kk < 0) {
            return false;
        }
    }
    return true;
}

signed main(void) {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n, k;
    cin >> n >> k;
    vector<int> v(n);
    for(int i = 0; i < n; i++) {
        cin >> v[i];
    }
    int l = 1, r = 1e9;
    int ans = 1e9;
    while(l <= r) {
        int mid = (l + r) / 2;
        if(check(mid, v, k)) {
            ans = min(ans, mid);
            r = mid - 1;
        }
        else {
            l = mid + 1;
        }
    }         
    cout << ans << endl;
}