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)

#pragma region RegionDefs
#include <bits/stdc++.h>
#define rep(i,n) for(int i=0,i##_len=(n);i<i##_len;++i)
#define reps(i,l,r) for(int i=(l),i##_len=(r);i<i##_len;++i)
#define repr(i,l,r) for(int i=(r)-1,i##_left=(l);i>=i##_left;--i)
#define all(x) begin(x),end(x)
using namespace std;
typedef long long ll;
const ll INF = 1e9;
template<class T=ll> using V = vector<T>;
template<class T=ll> using PQ = priority_queue<T>;
template<class T=ll> using PQG = priority_queue<T, V<T>, greater<T>>;
const ll MOD = 1000000007LL;
void in() {}
template<class Head, class... Tail>
void in(Head&& head, Tail&&... tail) { cin >> head; in(move(tail)...); }
#define IN(...) ll __VA_ARGS__; in(__VA_ARGS__)
#define INS(T, ...) T __VA_ARGS__; in(__VA_ARGS__)
#define VIN(T, v, n) V<T> v(n); for(auto& _elem:v)cin>>_elem
template<class T>bool chmax(T &a, const T &b) {if (a<b) {a=b; return 1;} return 0;}
template<class T>bool chmin(T &a, const T &b) {if (a>b) {a=b; return 1;} return 0;}
template<class T>
string join(vector<T>& v, string delim="") { ostringstream os; rep(i,v.size())i?os<<delim<<v[i]:os<<v[0]; return os.str(); }
#pragma endregion RegionDefs

// index が条件を満たすかどうか
bool f(vector<ll>& a, int len, int k)
{
    rep(i, a.size()) {
        k -= (a[i]+len-1)/len-1;
    }
    return k >= 0;
}

// 汎用的な二分探索のテンプレ。[s, t)からf(x)=trueとなる最小のxを探す
int meguru_search(vector<ll>& a, int s, int t, int k)
{
    int ng = s-1;
    int ok = t;
    while (abs(ok - ng) > 1)
    {
        int mid = (ok + ng) / 2;
        if (f(a, mid, k)) ok = mid;
        else ng = mid;
    }
    //okがtなら全ての要素でfalse
    //okがsなら全ての要素でtrue
    return ok;
}

void solve()
{
    IN(n, k);
    VIN(ll, a, n);
    ll res = meguru_search(a, 1, 1000000009LL, k);
    cout << res << endl;
}

int main()
{
    cin.tie(0); cout.tie(0);
    ios::sync_with_stdio(false);
    cout << setprecision(numeric_limits<double>::max_digits10);

    solve();

    return 0;
}