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;
#define fio ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL);
#define ll long long
#define dd double
#define pb push_back
#define ff first
#define ss second
#define Mp make_pair
const ll Mod=1000000007;
const ll INF=999999999999999999;
const ll NN=(ll)(1e6+5);
ll min(ll x,ll y){if(x<y) return x;return y;}
ll max(ll x,ll y){if(x>y) return x;return y;}
ll power(ll x,unsigned ll y) 
{ 
    if (y==0) 
        return 1; 
    else if (y%2==0) 
        return power(x,y/2)*power(x,y/2); 
    else
        return x*power(x,y/2)*power(x,y/2); 
}
ll hcf(ll a,ll b) 
{ 
    if(b==0) 
        return a; 
    return hcf(b,a%b);  
}
ll n,k;
vector <ll> v;
ll check(ll x)
{
    ll ans=0;
    for(auto u:v)
     ans+=(u-1)/x;
    return ans<=k; 
}
int main()
{
   fio;
   /*#ifndef ONLINE_JUDGE 
    	freopen("input.txt", "r", stdin); 
    	freopen("output.txt", "w", stdout); 
    	freopen("error.txt", "w", stderr); 
   #endif*/
   ll TT=1;
   //cin>>TT;
   while(TT--)
   {
       cin>>n>>k;
       v.resize(n);
       for(auto &u:v)
        cin>>u;
       ll l=1,r=1e9,mid;
       while(l<r)
       {
           mid=(l+r)/2;
           if(check(mid))
            r=mid;
           else
            l=mid+1; 
       } 
       cout<<l;
   }
}