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 rep(i, n) for(int i=0; i<n; i++)
#define rep1(i, n) for(int i=1; i<=n; i++)
#define repr(i, n) for(int i=n-1; i>=0; i--)
#define repr1(i, n) for(int i=n; i>=1; i--)
#define all(v) v.begin(),v.end()
using ll = long long;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
const int INF = 1e9;
const ll LLINF = 1e18;
const ll MOD = 1e9+7;
const double EPS = 1e-10;
const double PI = acos(-1);
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

int main() {
    int n; cin >> n;
    ll k; cin >> k;
    vector<ll> a(n);
    rep(i, n) cin >> a[i];

    if (k == 0) {
        ll m = -1;
        rep(i, n) chmax(m, a[i]);
        cout << m << endl;
        return 0;
    }

    ll ng = 0, ok = 2e18;
    while (ok - ng > 1) {
        ll mid = (ok + ng) / 2;
        ll sum = 0;
        rep(i, n) {
            if (a[i] <= mid) continue;
            sum += a[i] / mid;
        }
        if (sum > k) ng = mid; // 短く切りすぎ
        else ok = mid;
    }
    cout << ok << endl;

}