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;
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define repr(i,n) for(int i = (int)(n); i >= 0; i--)
#define all(v) v.begin(),v.end()
typedef long long ll;

ll min_cut(vector<ll> &A, ll x, ll N){
    ll ans = 0;
    rep(i,N){
        if (A[i] % x == 0) ans += (A[i] / x - 1);
        else ans += (A[i] / x);
    }
    return ans;
}

int main(){
    ll N,K;
    cin >> N >> K;
    vector<ll> A(N);
    rep(i,N){
        cin >> A[i];
    }
    sort(all(A));
    ll ans = 0;
    if (min_cut(A, 1, N) <= K){
        ans = 1;
    }
    else{
        ll left = 1;
        ll right = A[N - 1];
        while(true){
            if (right - left == 1){
                ans = right;
                break;
            }
            else{
                if (min_cut(A, (left + right) / 2, N) <= K){
                    right = (left + right) / 2;
                }
                else{
                    left = (left + right) / 2;
                }
            }
        }
    }
    cout << ans << endl;
}