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>
#define REP(i,n) for(int i=0;i<(n);++i)
using namespace std;
typedef long long ll;

ll N,K;
vector< ll > A(200000);

// 各 Ai を a で割っていって、K 個以下に分割できたら true
// K 個より分割が多くなるなら false
bool check(ll k){
  ll divn = 0;
  REP(i,N) {
    divn += (A[i]-1) / k;
    if ( divn > K ) return false;
  }
  return true;
}

int main(){
  cin >> N >> K;
  REP(i,N) {
    cin >> A[i];
    A[i] *= 100;
  }
  ll l = 1;
  ll r = 100000000000;
  while ( l+1 < r ) {
    ll mid = (l + r) / 2;
//    cout << l << " " << mid << " " << r << " : ";
    if ( check( mid ) )
      r = mid;
    else
      l = mid;
//    cout << " -> " << l << " " << r << endl;
  }
  cout << ((r+99)/100) << endl;
}