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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define mp make_pair
#define pb(x) push_back(x)
#define vi vector<int>
#define vs vector<string>
#define vll vector<long long>
#define vc vector<char>
#define vs vector<string>
#define pii pair<int, int>
#define pll pair<long long, long long>
#define pil pair<int, long long>
#define pli pair<long long, int>
#define pcc pair<char, char>
#define pdd pair<double, double>
#define mll map<long long, long long>
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
#define flsh fflush(stdout)
#define gcd __gcd
#define clr(x) memset(x, 0, sizeof(x))
#define mod 1000000007LL
#define mod2 998244353LL
#define INF 1000000000007

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> o_tree;


ll check(ll a[],ll n,ll k,ll mid)
{
    ll i,j;
    ll res=0;
    for(i=0;i<n;i++)
    {
        res+=(a[i]/mid); 
    }
    if(res<=k)
    {
        return 1;
    }
    return 0;
}

void solve()
{
    ll n,k;
    ll i,j;
    cin>>n>>k;
    ll a[n];
    for(i=0;i<n;i++)
    {
        cin>>a[i];
    }
    ll maxi = *max_element(a,a+n);
    
    ll l=1,r=maxi;
    ll ans=maxi;
    while(l<=r)
    {
        ll mid = l+(r-l)/2;
        
        if(check(a,n,k,mid))
        {
            ans=mid;
            r=mid-1;
        }
        else{
            l=mid+1;
        }
    }
    cout<<ans<<"\n";

    

}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    if (fopen("input.txt", "r"))
    {
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    }
    cout << setprecision(20);

    ll t = 1;

    // cin>>t;

    while (t--)
    {
        solve();
    }

    return 0;
}