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;
///////////////////////////////////////////
const long long int INF = 1LL<<60;
const long long int Mod = 1000000007;
using ll = long long int; using ci = const int;
using vi = vector<int>;  using Vi = vector<long long int>;
using P = pair<int, int>;  using PLL = pair<ll, ll>;
using matrix = vector<vector<ll>>;
#define pb(x) push_back(x)
#define mp(x,y) make_pair(x,y)
#define all(x) (x).begin(),(x).end()
#define rp(i,N) for(ll i = 0; i < (ll)N; i++)
#define repi(i,a,b) for(ll i = ll(a); i < ll(b); ++i)
template<class T>bool chmax(T &former, const T &b) { if (former<b) { former=b; return true; } return false; }
template<class T>bool chmin(T &former, const T &b) { if (b<former) { former=b; return true; } return false; }
template<class T>T sqar(T x){ return x*x; }//sqrt(x)は平方根;
#define Sort(v) std::sort(v.begin(), v.end(), std::greater<decltype(v[0])>()) //降順でVをソート
#define p_queue(v) priority_queue<v, vector<v>, greater<v> >
template<class T> inline void princ(T x){cout<<x<<" ";}; 
template<class T> inline void print(T x){cout<<x<<"\n";};
template<class T> inline void Yes(T condition){ if(condition) cout << "Yes" << endl; else cout << "No" << endl; }
template<class T> inline void YES(T condition){ if(condition) cout << "YES" << endl; else cout << "NO" << endl; }
///////////////////////////////////////////////////////////////////////////////////

void solve(){
    ll n,k;
    cin >> n >> k;
    Vi a(n);
    rp(i,n) cin >> a[i];
    ll mid, ok = 1000000002, ng = 0;
    while(abs(ok-ng)>1){
        mid = (ok+ng)/2;
        ll ctr=0;
        rp(i,n) ctr+=(a[i]-1)/mid;
        if(ctr<=k)ok = mid;
        else ng = mid;
    }//midではなくokを使わないとバグる!
    print(ok);
    return;
}
int main(){
    cin.tie(0);ios::sync_with_stdio(false);
    std::cout<<std::fixed<<std::setprecision(30);
    solve();
    return 0;
}