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>
#define itn int
#define REP(i, n) for (ll i = 0; i < n; i++)
#define IREP(i, n) for (ll i = n - 1; i >= 0; i--)
#define FOR(i, a, b) for (ll i = a; i < b + 1; i++)
#define all(v) v.begin(), v.end()
#define SENTINEL 2000000000
#define NIL -1
using namespace std;
typedef long long ll;

const ll MAX = 510000;
const ll INF = 1LL << 60;
const ll MOD = 1000000007;
template <class T>inline bool chmin(T &a, T b){if(a>b){a=b;return true;}return false;}
template <class T>inline bool chmax(T &a, T b){if(a<b){a=b;return true;}return false;}


int main()
{
    int n;double k;
    cin >> n >> k;
    ll max=0;
    vector<ll> a(n);
    REP(i,n){
        cin >> a[i];
        chmax(max,a[i]);
    }
    ll l=0;
    ll ans=max/2;
    ll r=max;
    while(l<r){
        int kk=k;
        REP(i,n){
            if(a[i]<=ans){
                continue;
            }
            int tmp=a[i]/ans;
            if(a[i]%ans==0){
                tmp--;
            }
            kk-=tmp;
        }
        if(kk>=0){
            l=ans;
            ans=(l+r)/2;
        }
        else{
            r=ans+1;
            ans=(l+r)/2;
        }
    }
    cout << r << endl;
}