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;
#define ll long long
#define pb push_back
#define mp make_pair
#define mt make_tuple
#define ff first
#define ss second
#define pii pair <int,int>
#define pll pair <ll,ll>
#define testcase() int t; cin>>t; while(t--)
#define forn(i,n) for(int i=0;i<n;i++)
#define forn1(i,n) for(int i=1;i<=n;i++)
#define vll vector <ll>
#define vi vector <int>
#define all(v) v.begin(),v.end()

ll M = 1e9 + 7;
double pi = acos(-1.0);
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
ll powerm(ll x,ll y){ ll res=1; while(y){ if(y&1) res=(res*x)%M; y=y>>1; x=(x*x)%M;} return res%M; }
ll power(ll x,ll y){ ll res=1; while(y){ if(y&1) res=(res*x); y=y>>1; x=(x*x);} return res; }
ll gcd(ll a,ll b){if(b>a)return gcd(b,a); if(b==0)return a; return gcd(b,a%b);}

int main() {
	
	//ifstream cin("input.txt");
	//ofstream cout("output.txt");
	ios_base::sync_with_stdio(false); 
    cin.tie(NULL); 
	
	int n; ll k;
    cin>>n>>k;
    vll a(n);
    forn(i,n) cin>>a[i];
    if(k==0)
    {
        ll mx = 0;
        forn(i,n) mx = max(mx,a[i]);
        cout<<mx; return 0;
    }
    ll L = 1, R = 1e9+1;
    while(R-L>1)
    {
        ll mid = (L+R)/2;
        ll c = 0;
        forn(i,n)
        {
            ll g = (a[i]-1)/mid;
            c += g;
        }
        //cout<<mid<<' '<<c<<endl;
        if(c>k) L = mid;
        else R = mid;
    }
    cout<<R;

	return 0;
}