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>

typedef uint64_t u64;
typedef int64_t i64;
typedef uint32_t u32;
typedef int32_t i32;
typedef uint16_t u16;
typedef int16_t i16;
typedef uint8_t u8;
typedef int8_t i8;

using namespace std;

#define rep(idx, N) for (u64 idx = 0; idx < N; idx++)

int main() {
  u64 n, k;
  cin >> n >> k;

  vector<u64> a(n);
  u64 max = 0;
  rep(i, n) {
    cin >> a[i];
    if (a[i] > max)
      max = a[i];
  }

  u64 r = 0;
  u64 l = max;
  u64 cnt = 0;

  while (r + 1 != l && cnt < 100000000) {
    u64 m = (r + l) / 2;
    u64 ki = 0;
    rep(i, n) {
      if (a[i] > m) {
        ki += (a[i] - 1) / m;
      }
    }

    if (ki <= k) {
      l = m;
    } else {
      r = m;
    }
    // cout << "l:" << l << "\tr:" << r << endl;
    cnt++;
  }

  cout << r + 1;

  return 0;
}