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>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;

#define all(v) v.begin(), v.end()

const ll mod=1000000007;

ll countBits(ll in){
  int res=0;
  for(;in>0;in>>=1){
    if((in&0x01)!=0){
      res++;
    }
  }
  return res;
}

template<typename T>
void show2dVector(vector<vector<T>>& v){
  for(int i=0;i<v.size();i++){
    int m=v[i].size();
    cout<<i<<" : ";
    for(int j=0;i<m;i++){
      cout<<v[i][j]<<" ";
    }
    cout<<endl;
  }
}

void bfs(const vector<vector<int>>&g, int v, vector<bool>& seen){
  seen[v]=true;
  cout<<v<<" ";
  for(auto next: g[v]){
    if(seen[next]){
      continue;
    }
    bfs(g,next,seen);
  }
}

bool dfs(vector<vector<int>> &g,int start,int goal,vector<bool>&seen){
  bool res=false;
  seen[start]=true;
  if(start==goal){
    return true;
  }
  for(auto next: g[start]){
    if(seen[next]){
      continue;
    }
    res=dfs(g,next,goal,seen);
    if(res){
      break;
    }
  }
  return res;
}

ll gcd(ll a,ll b) {
  return b? gcd(b,a%b): a;  
}
ll lcm(ll a,ll b) {
  return a*b/gcd(a,b);
}
bool isLowerCase(char c){
  return (c>='a'&&c<='z');
}

ll powm(ll a,ll n, ll m){
  ll ret=1;
  while(n>0){
    if(n%2==1){
      ret=(ret*a)%m;
    }
    n>>=1;
    a=(a*a)%m;
  }
  return ret;
}

void primeFactor(ll val, map<ll,ll> &primeList){
  if(val<1){
    return;
  }
  for(int i=2;i*i<val;i++){
    while(val%i==0){
      primeList[i]++;
      val/=i;
    }
  }
  if(val!=1){
    primeList[val]=1;
  }
  return;
}

const string yesno(bool ans){
  return (ans?"Yes":"No");
}
int main() {
  int n,k;cin>>n>>k;
  vector<int> a(n);
  for(int i=0;i<n;i++){
    cin>>a[i];
  }
  int l,r,x;
  l=0;r=INT_MAX;
  while(l+1<r){
    x=(l+r)/2;
    int cnt=0;
    for(int i=0;i<n;i++){
      int d=(a[i]+x-1)/x-1;
      if(d>0){
        cnt+=d;
      }
    }
    if(cnt<=k){
      r=x;
    }else{
      l=x;
    }
  }
  cout<<x<<endl;
  return 0;
}