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>

const int MAX_N = 200005;
int N, K;
int logs[MAX_N];

bool check(int maxlen) {
  if (maxlen == 0) return false;
  int k = 0;
  for (int i = N -1; i  >= 0; --i) {
    int log = logs[i];
    if (log <= maxlen) continue;
    k += log / maxlen - !(log % maxlen);
    if (k > K) return false;
  }
  return true;
}

int main() {
  scanf("%d %d", &N, &K);
  int mm = 0;

  for (int i = 0; i < N; ++i) {
    scanf("%d", logs + i);
    mm = std::max(mm, logs[i]);
  }
  std::sort(logs, logs + N);

  long long hi = 1ll<<31;
  long long lo = 0;
  while (hi - lo > 1) {
    long long mid = (hi + lo) / 2;
    if (check(mid)) {
      hi = mid;
    } else {
      lo = mid;
    }
  }
  if (check(lo)) printf("%lld\n", lo);
  else printf("%lld\n", hi);
}