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 ll long long int
#define vi vector<int>
#define vll vector<ll>
#define vvi vector < vi >
#define pii pair<int,int>
#define pll pair<long long, long long>
#define vpll vector < pair<long long,long long> >
#define fill(a,val)  memset(a,val, sizeof (a))
#define sort_unique(c) (sort(c.begin(),c.end()), c.resize(distance(c.begin(),unique(c.begin(),c.end()))))
#define mp make_pair
#define pb push_back
#define f first
#define s second
#define all(cont) cont.begin(), cont.end()
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
#define MAX   1e9
#define MIN  -1e9
typedef map<int,int> mpi;
typedef set<int> seti;
typedef multiset<int> mseti;
#define fio ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define rep(i, n)    for(long long int i = 0; i < (n); ++i)
#define repA(i, a, n)  for(int i = a; i <= (n); ++i)
#define repD(i, a, n)  for(int i = a; i >= (n); --i)
#define repn(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
#define trav(a, x) for(auto& a : x)
struct Interval{ll start,end;};
bool compare(Interval i1, Interval i2) {return (i1.start < i2.start);}

vll divisors[10000001];
int main(){
	fio;
	#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	#endif
	ll n,k;
	cin>>n>>k;
	ll a[n];
	rep(i,n){
		cin>>a[i];
	}
	sort(a,a+n);
	ll l = 1,r = a[n-1];
	ll ans = 10000000000;
	while(l <= r){
		ll mid = (l+r)/2;
		ll total = 0;
		rep(i,n){
			if(mid<a[i]){
				if(a[i]%mid == 0){
					total+=a[i]/mid - 1;
				}
				else{
					total+=a[i]/mid;
				}
			}
		}
		if(total<=k){
			ans = min(ans,mid);
			r = mid-1;
		}
		else{
			l = mid+1;
		}
	}
	cout<<ans;
	return 0;
}