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: Wrong Answer)

#include <bits/stdc++.h>
#define rup(x,y) (((x)+(y)-1)/(y))
using namespace std;
using ll = long long;
using P = pair<int,int>;

int main() {
    int n, k;
    cin >> n >> k;
    vector<int> a(n);
    int l = 0, r = 0;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        r = max(r, a[i]);
    }
    
    int m = 0, ans = r;    
    while (l < r) {
        m = (l + r) / 2;
        if (m < 1) {
            ans = 1;
            break;
        }
        int count = 0;
        for (int i = 0; i < n; i++) {
            count += rup(a[i], m) - 1;
        }
        if (count > k)
            l = m + 1;
        if (count <= k) {
            ans = m;
            r = m - 1;
        }
    }
    while (1) {
        if (m < 1) {
            ans = 1;
            break;
        }
        int count = 0;
        for (int i = 0; i < n; i++) {
            count += rup(a[i], m) - 1;
        }
        if (count <= k) {
            ans = m;
            m--;
        }
        else
            break;
    }
    
    cout << ans << endl;
    
    return 0;
}