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>

using namespace std;
using ll = long long;
using vl = vector<ll>;
using pll = pair<ll, ll>;

#define rep(i, n) for (ll i = 0, xxx_rep_end = (n); i < xxx_rep_end; ++i)
#define all(x) (x).begin(), (x).end()

template <class T1, class T2>
inline void chmax(T1 &a, const T2 &b)
{
    if (a < b)
    {
        a = b;
    }
}
template <class T1, class T2>
inline void chmin(T1 &a, const T2 &b)
{
    if (a > b)
    {
        a = b;
    }
}

constexpr ll INF = 0x3f3f3f3f3f3f3f3fLL;

inline bool C(ll num, ll K, const vl &A)
{
    ll sum = 0;
    for (auto a : A)
    {
        sum += a / num;
        if (sum > K)
        {
            return false;
        }
    }
    return true;
}

int main()
{
    ll N, K;
    cin >> N >> K;
    vl A(N);
    rep(i, N) cin >> A[i];
    sort(all(A), greater<ll>());

    ll lb = 1;
    ll ub = A[0] + 1;
    while (ub - lb > 1)
    {
        ll mid = (lb + ub) / 2;
        if (C(mid, K, A))
        {
            ub = mid;
        }
        else
        {
            lb = mid;
        }
    }
    cout << ub << endl;
}