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: Wrong Answer)

#include<bits/stdc++.h>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
using namespace __gnu_pbds;
#define ll long long
#define ld long double
#define DEBUG(x) cout << '>' << #x << ':' << x << endl;
#define REP(i,n) for(ll i=0;i<(n);i++)
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define FORC(i,a,b,c) for(ll i=(a);i<(b);i+=(c))
#define pb(x) push_back(x)
#define ff first
#define ss second
#define dd double
#define mod 100000000000000007

template<class K,class other=null_type,class cmp=less<K>> using ordered_set=tree<K,other,
cmp, rb_tree_tag, tree_order_statistics_node_update>;

inline ll md(ll x){
    return x<mod?x:x%mod;
}

ll power(ll x, ll y){
    if(y==0){
        return 1;
    }
    ll x_=power(x,y/2);
    if(y & 1){
        return md(md(x_*x_)*md(x));
    }
    else{
        return md(x_*x_);
    }
}

ll mod_inverse(ll x){
    return power(x,mod-2);
}
ll n,k;
const ll N=2e5+10;
ll arr[N];
bool check(ll tar){
    if(tar==0){
        ll sum=0;
        for(ll i=0;i<n;i++){
            sum+=2*arr[i]-1;
        }
        if(sum<=k){
            return true;
        }
        else{
            return false;
        }
    }
    ll sum=0;
    for(ll i=0;i<n;i++){
        if(arr[i]<=tar){
            continue;
        }
        ll x=arr[i]/tar;
        ll y=round(double(arr[i])/double(x));
        sum+=(x-1);
        if(y!=tar){
            sum++;
        }
    }
    if(sum<=k){
        return true;
    }
    else{
        return false;
    }
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin>>n>>k;
    REP(i,n){
        cin>>arr[i];
    }
    ll l=0,h=1e9;
    while(l!=h){
        ll m=(l+h)>>1;
        if(check(m)){
            h=m;
        }
        else{
            l=m+1;
        }
    }
    cout<<l<<endl;

}