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<iostream>
#include<vector>
#include<cstring>
#include<cmath>
#include<set>
#include<queue>
#include<algorithm>
#include<climits>
#include<map>
#include<unordered_map>
#define ll long long

bool isPrime(ll n){
    if(n%2==0 || n%3==0) return false;
    if(n<=1)  return false;
    if(n<=3)  return true;
    for(ll i=5;i*i<=n;i=i+6)
        if(n%i==0||n%(i+2)==0)
           return false;
    return true;
}

#define w(t) ll t,tt=1;	cin>>t;		while(t--)
#define ps(x,y) fixed<<setprecision(y)<<x
#define setbits(x) __builtin_popcountll(x)
#define zrobits(x) __builtin_ctzll(x)
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl "\n"
#define ff first
#define ss second
#define pb push_back
#define mp make_pair

#define vi vector<int>
#define vll vector<ll>
#define mii map<ll,ll>
#define umii unordered_map<ll,ll>
#define pi pair<int,int>
#define pll pair<ll,ll>
#define vpii vector<pair<int,int>>
#define vpll vector<pair<ll,ll>>

#define mod 1e9+7
#define inf 1e18

#define lln ll n; cin>>n
#define intn int n; cin>>n
#define YES cout<<"YES"
#define NO cout<<"NO"
#define strr    string str; cin>>str;
#define strn   n=str.length()
#define fr(a,n) for(ll i=a;i<n;i++)
#define loop for(ll i=0;i<n;i++)

using namespace std;

bool possible(int mid,int k,int arr[],int n){
    if(mid==0)  return false;
    int curr=0;
    loop{
        if(arr[i]>mid)   curr+=(arr[i]/mid);
    }
    return curr<=k;
}

void solve (){
    intn;
    int k;
    cin>>k;
    int arr[n];
    loop cin>>arr[i];
    int mid,low=1,high=10;
    while(low<high){
        mid=(low+high)/2;
        //cout<<low<<" "<<high<<" "<<mid<<endl;
        if(possible(mid,k,arr,n)) high=mid-1;
        else                low=mid+1;
    }
    //cout<<low<<" "<<high<<" "<<mid<<endl;
    cout<<high;
}

int main(){
    IOS
    solve();
    return 0;
}