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;
typedef long long ll;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define agewari(a, b) ((ll)a + ((ll)b - 1)) / b
const int MOD = 1000000007;
const long long INF = 1LL << 60;
using Graph = vector<vector<ll>>;

ll n, k;
vector<ll> A;

int ok(ll x)
{
    ll cnt = 0;
    rep(i, n)
    {
        cnt += agewari(A[i], x) - 1;
    }
    if (cnt <= k)
        return 1;
    else
        return 0;
}

int main()
{
    cin >> n >> k;
    ll Amax = 0;
    rep(i, n)
    {
        ll a;
        cin >> a;
        Amax = max(Amax, a);
        A.push_back(a);
    }

    //にぶたん
    ll left = 0;
    ll right = Amax + 1;
    while (right - left > 1)
    {
        ll mid = left + (right - left) / 2;
        if (ok(mid))
            right = mid;
        else
            left = mid;
    }

    

    cout << right << endl;
}