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: Time Limit Exceeded)

#include <bits/stdc++.h>

using namespace std;

#define bug(x) cerr << #x << " = " << x << '\n'
#define ll     long long
#define maxn   200005

int n, k;
int a[maxn];

bool f (double x) {
        multiset<int> s;
        for(int i = 1; i <= n; ++ i)
            s.insert(a[i]);
        for(int i = 1; i <= k; ++ i) {
            int u = *s.rbegin();
            if(u > x) {
                s.erase(s.find(u));
                s.insert(x); s.insert(u - x);
            }
        }
        return *s.rbegin() <= x;
}

int main() {
        //freopen(".inp","r",stdin);

        cin >> n >> k;

        for(int i = 1; i <= n; ++ i)
            cin >> a[i];

        int lo = 0, hi = *max_element(a + 1, a + n + 1), ret = hi;

        while(hi >= lo) {
            int mid = (hi + lo) / 2;
            if(f(mid)) ret = mid, hi = mid - 1;
            else lo = mid + 1;
        }

        cout << ret;
}