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: Runtime Error)

#include <bits/stdc++.h>
#include <chrono>
#include <math.h>
using namespace std;
using ll = long long;
using ld = long double;
using namespace std::chrono; 
#define fo(i,z,n) for(int i = z;i<n;i++)
#define Fo(i,z,n) for(ll i = z;i<n;i++)
#define INT_BITS 32
#define mod 1000000007
#define vi vector<int>
#define vll vector<ll>
#define vvi vector<vector<int> >
#define vvl vector<vector<ll> > 
#define pb push_back
#define f first
// #define s second
#define int long long
#define all(c) (c).begin(), (c).end()
#define endl '\n'
// auto start = high_resolution_clock::now(); 
// auto stop = high_resolution_clock::now(); 
// auto duration = duration_cast<microseconds>(stop - start); 
 
// cout << duration.count() << endl; 
int ceil(int a,int b){
	if(a%b == 0)return a/b;
	else return a/b+1;
}


bool check(int* arr,int mid ,int k,int n){
	int r = 0;
	fo(i,0,n){
		r += ceil(arr[i],mid)-1;
	}
	if(r <= k)return true;
	else return false;
}



int32_t main()
{
 
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
     cout.tie(0);
     // int t;
     // cin >> t;

     // while(t--){

 
     // }

     int n,k;
     cin >> n >> k;
     int arr[n];
     fo(i,0,n)cin >> arr[i];
sort(arr,arr+n);

int l = 0;
int r = mod;
int ans = -1;
while(l < r){
	int mid = (l+r)/2;
	// cout << mid << endl;
	if(check(arr,mid,k,n)){
		ans = mid;
		r = mid;
	}
	else l = mid+1;
}

cout << ans << endl;

     

return 0;
}