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 <algorithm>
#include <iostream>
#include <iomanip>
#include <cassert>
#include <cstring>
#include <string>
#include <vector>
#include <random>
#include <bitset>
#include <queue>
#include <cmath>
#include <unordered_map>
#include <set>
#include <map>
#define INCANT cin.tie(0), cout.tie(0), ios::sync_with_stdio(0), cout << fixed << setprecision(20);
#define rep(i,n) for (int i=0; i<n;++i)
#define ALL(a)  (a).begin(),(a).end()
typedef long long ll;
using namespace std;
const ll MOD = 1e9+7LL;
const int INF = 2e9;
int N, K;
int A[200003];

bool is_large(int X) {
  int cnt=0;
  rep(i, N) {
    cnt+=(A[i]+X-1)/X-1;
  }
  return cnt > K;
}

int binary_search(int l, int r) {
  while(r-l>1) {
    int mid = (r+l)/2;
    if (is_large(mid)) l = mid;
    else r = mid;
  }
  return r;
}

int main() {
  INCANT;
  cin >> N >> K;
  rep(i, N) cin >>A[i];
  cout << binary_search(0, 1e9) << endl;

  return 0;
}