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++ (Clang 10.0.0) (judge verdict: Accepted)

#include <iostream>
#include <vector>
#include <string>
#include <numeric>
#include <tuple>
#include <algorithm>
#include <array>
using namespace std;
using ll = long long;
#define rep(i,n) for(decltype(n) i=0; i<n; ++i)
inline constexpr ll Inf = (1ULL << 62) -1;

template <class T>
void updatemax(T& a, T b) { if (b > a) a = b; }

int main() {
    ll N, K;
    cin >> N >> K;
    vector<ll> A(N);
    rep(i,N) cin >> A[i];
    ll l = 0;
    ll r = 1e9+1;
    auto f = [&](auto len) {
        ll cut = 0;
        rep(i,N) {
            cut += (A[i]-1) / len;
        }
        return cut <= K;
    };
    while (r-l > 1) {
        ll m = (r + l)/2;
        if (f(m)) r = m;
        else l = m;
    }
    ll ans=r;
    cout << ans << endl;
    return 0;
}