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>
using namespace std;

const int MOD = 1000000007;
const int INF = 1e9;

int main(){
  int n,k;
  const int max_n = 2 * 1e5;
  int a[max_n + 1];
  cin >> n >> k;
  int max_a = 0;
  for(int i = 0; i < n; i++) {
    cin >> a[i];
    max_a = max(max_a, a[i]);
  }
  int left = 0;
  int right = max_a;
  int x;
  int sum = 0;
  while(right - left > 1){
    sum = 0;
    x = (right + left) / 2;
    for(int i = 0; i < n; i++){
      if(a[i] == x) sum += a[i] / x - 1;
      else sum += a[i] / x;
    }
    // cout << "x= " << x << ", sum = " << sum << endl;
    if(sum <= k) right = x;
    else left = x;
    // if(right - left <= 1) break;
  }

  cout << right << endl;

  return 0;
}