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 maxN 200005
#define inf (int) 2e9
#define FOR(i, a, b) for(int i = a; i <= b; i++)
#define REP(i, a, b) for(int i = a; i >= b; i--)

using namespace std;

int n, k;
int a[maxN];

void enter() {
    cin >> n >> k;
    FOR(i, 1, n)
       cin >> a[i];
    sort(a + 1, a + n + 1);
}

bool ok(int x) {
    int ans = 0;
    FOR(i, 1, n) {
        int y = a[i], u = y / x;
        while(x * u < y) u++;
        ans += u - 1;
    }
    return ans <= k;
}

void solve() {
    int lo = 1, hi = inf;
    while(hi - lo > 1) {
        int m = (hi + lo)/2;
        if(ok(m)) hi = m;
        else lo = m;
    }
    cout << hi;
}

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