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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
#define ll long long
void A(){
    int x;
    cin>>x;
    if(x>=30) cout<<"Yes"<<endl;
    else cout<<"No"<<endl;
}
void B(){
    ll n,d;
    cin>>n>>d;
    int res=0;
    for(int i=0;i<n;i++){
        ll x,y;
        cin>>x>>y;
        if(x*x+y*y<=d*d) res++;
    }
    cout<<res<<endl;
}
void C(){

}
void D(){
    int n;
    string s;
    cin>>n;
    cin>>s;
    stack<int>st;
    for(int i=0;i<n;i++){
        if(s[i]=='R') st.push(i);
    }
    int l=0,r=n-1;
    int res=0;
    for(int i=0;i<n&&!st.empty();i++){
        if(i>=st.top()) break;
        if(s[i]=='W'){
            st.pop();
            res++;
        }
    }
    cout<<res<<endl;
}
void E(){
    ll n,k;
    cin>>n>>k;
    vector<ll> a(n,0);
    ll ma=0;
    for(int i=0;i<n;i++){
        cin>>a[i];
        ma=max(ma,a[i]);
    }
    if(k==0) {
        cout<<ma<<endl;
        return;
    }
    ll l=0,r=ma;
    ll ans=0;
    while(l<=r){
        ll mid=l+(r-l)/2;
        ll cnt=0;
        for(ll i:a){
            cnt+=(i-1)/mid;
        }
        if(cnt<=k) {
            ans=mid;
            r=mid-1;
        }
        else l=mid+1;
    }
    cout<<ans<<endl;
}
void F(){

}
void helper(){
    ll r=1;
    while(r<LONG_LONG_MAX/100){
        vector<ll> res;
        for(ll i=1;i*i<=r;i++){
            if(r%i==0){
                res.push_back(i);
                if(r/i!=i) res.push_back(r/i);
            }
        }
        sort(res.begin(),res.end());
        for(ll i:res) cout<<i<<" ";
        cout<<endl;
        r=r*10+1;
    }
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
//    helper();
    E();
    return 0;
}