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 <algorithm>
#include <cmath>
#include <cstdio>
#include <deque>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>

using namespace std;

typedef long long ll;

#define MOD 1000000007

int n;
ll k;
vector<ll> a;

bool check(ll x) {
  ll cnt = 0;
  for (int i = 0; i < n; ++i) {
    cnt += a[i] / x;
  }
  return cnt <= k;
}

int main() {
  cin >> n >> k;
  a.resize(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  ll ok = a[n - 1], ng = 0;
  while (1 < abs(ok - ng)) {
    ll mid = (ok + ng) / 2;
    if (check(mid)) {
      ok = mid;
    } else {
      ng = mid;
    }
  }
  cout << ok << endl;
  return 0;
}