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)

//FIRST THINK THEN CODE.

#include <bits/stdc++.h>
 
 
using namespace std;
 
typedef long long ll;
 
#define rep(i,a,b) for(ll i=a;i<b;++i)
#define rrep(i,a,b) for(ll i=a;i>b;--i)
#define FOR(i,n)  for(ll i=0;i<n;i++)
#define vi vector<int>
#define vl vector<ll>
#define ld long double
#define vvi vector<vector<int>>
#define vvl vector<vector<long long>>
#define pii pair<int,int>
#define pll pair<long,long>
#define vpii vector<pii>
#define vpll vector<pll>
#define ff first
#define ss second
#define pb push_back
#define pf push_front
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define bs binary_search
#define d1(x) cout<<(x)<<endl
#define d2(x,y) cout<<(x)<<" "<<(y)<<endl
#define d3(x,y,z) cout<<(x)<<" "<<(y)<<" "<<(z)<<endl
#define d4(a,b,c,d) cout<<(a)<<" "<<(b)<<" "<<(c)<<" "<<(d)<<endl
#define PI 3.1415926535897932384626433832795
#define fix(f,n) fixed<<setprecision(n)<<f
#define all(x) x.begin(),x.end()
#define rev(p) reverse(p.begin(),p.end());
#define endl "\n"
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define popcount(x) __builtin_popcountll(x)
#define sz(x) ((ll)x.size())
const ll M=1000000007;
const ll MM=998244353;
ll begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";

 
ll newmod(ll a,ll b)
 {
  return ((a%b)+b)%b;
}
 
 
ll powM(ll a,ll b,ll m )
{ 
  if(a<=0)return 0;
  a%=m;
  if(!b)return 1;
  ll ans=1LL;
  while(b)
  {
    if(b&1)ans=ans*a%m;
    a=a*a%m;
    b>>=1;
  }
 
return ans;
}


 
ll poww(ll a,ll b)
{ 
  
  ll ans=1;
  while(b)
  {
    if(b&1)ans=ans*a;
    a=a*a;
    b>>=1;
  }
 
return ans;

}

template<typename T,typename F>
void chmax( T &a,F b){
  if(b>a)a=b;
}

template<typename T,typename F>
void chmin( T &a,F b){
  if(b<a)a=b;
}



const ll N=2e5+5;


ll f(ld x){
  ld rem=x-floor(x);
  if(rem>=0.5)return (ll)ceil(x);
  else return (ll)floor(x);
}

int main()
{ 

IOS;


ll n;cin>>n;
ll k;cin>>k;
vector<ld> a(n);

FOR(i,n)cin>>a[i];

ld l=0,r=1e18,m;

ll ans=1e18;

ll cnt=0;

//  ll s=0;

//cout<<log2(4.6);

//
//d1(s);

//cout<<f(3.5)<<endl;



while(r-l>0.001){
  cnt++;
  m=(l+r)/(2*1.0);

  ll s=0;
  FOR(i,n){
    ll temp=ceil(1.0*a[i]/m);
    s+=temp-1;
    if(s>k)break;
  }

//d3("hi",s,m);

  if(s>k){
    l=m;
  }
  else {
    ll t=round(m);
    chmin(ans,t);
    r=m;
  }

}

cout<<ans;


//end_routine();

return 0;
 
 }