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)

//header{{{
#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
//#include<boost/multiprecision/cpp_int.hpp>
//#include<boost/math/common_factor_rt.hpp>
  
using namespace std;
//using namespace boost::multiprecision;  
//using namespace boost::math;

#define rep(i,n) for(int i=0;i<(n);++i)
#define reps(i,n) for(int i=1;i<=(n);++i)
#define all(x) (x).begin(),(x).end()
#define Fixed fixed << setprecision(12)
#define int int64_t
using pii = pair<int,int>;
constexpr int INF  = 0x3f3f3f3f;
constexpr long long LINF = 0x3f3f3f3f3f3f3f3fLL;
constexpr int mod1 = 1e9+7; 
constexpr int mod2 = 998244353;

template <class A, class B> inline bool chmax(A &a, const B &b) { return b > a && (a = b, true); }
template <class A, class B> inline bool chmin(A &a, const B &b) { return b < a && (a = b, true); }

template <class T> using min_heap = priority_queue<T,vector<T>,greater<T> >;
template <class T> using max_heap = priority_queue<T>;
template <class A, class B> using umap = unordered_map<A,B>;
  
inline int square(int a){ return a * a;}
inline int updiv(int a,int b){ return (a + b - 1) / b; }

constexpr int dx[] = {1,0,-1,0,1,1,-1,-1};
constexpr int dy[] = {0,-1,0,1,1,-1,-1,1};
//}}}


signed main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n, k;
    cin >> n >> k;

    vector<int> a(n);

    rep(i,n){
        cin >> a[i];
    }

    auto f = [&](auto key){
        int cnt = 0;
        rep(i,n){
            cnt += updiv(a[i], key) - 1;
        }
        return (cnt);
    };

    int ok = 1e9 + 1, ng = 0;
    
    while(ok - ng > 1){
        auto mid = (ok + ng) / 2;
        if(f(mid) <= k){
            ok = mid;
        }else{
            ng = mid;
        }
    }

    cout << ok << '\n';

    return 0;
}