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 rep(i,n) for(int i=0;i<(int)(n);i++)
using namespace std;
using ll = long long ;
using P = pair<int,int> ;
using pll = pair<long long,long long>;
constexpr int INF = 1e9;
constexpr long long LINF = 1e17;
constexpr int MOD = 1000000007;
constexpr double PI = 3.14159265358979323846;
ll n;
ll k;
vector<ll> a(n);

bool ok(ll m){
    ll cnt = 0;
    rep(i,n){
        cnt += (a[i]+m-1)/m;
    }
    cnt -= n;
    return cnt <= k;
}

int main(){
    cin >> n >> k;
    a.resize(n);
    rep(i,n) cin >> a[i];

    ll left = -1,right = INF + 100;
    while(right - left > 1LL){
        ll mid = (right + left)/2;
        if(ok(mid)) right = mid;
        else left = mid;
    }
    cout << right << endl;
    return 0;
}