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;
typedef unsigned int uint;
typedef unsigned long long ull;
static const double EPS = 1e-12;
static const double PI = acos(-1.0);

template <class T>
using V = vector<T>;
template <class T>
using VV = V<V<T>>;

#define FOR(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
#define REP(i, n) for (int i = 0; i < (int)(n); i++)
#define ALL(a) (a).begin(), (a).end()
#ifdef LOCAL
#define dbg(x) cerr << __LINE__ << " : " << #x << " = " << (x) << endl
#else
#define dbg(x) true
#endif

// index が条件を満たすかどうか
template <typename T>
bool isOK(const vector<T> &a, ll index, T key) {
  REP(i, a.size()) {
    key -= (ll)ceil(double(a[i]) / double(index)) - 1;
  }
  if (key >= 0)
    return true;
  else
    return false;
}

// 汎用的な二分探索のテンプレ
template <typename T>
ll binary_search(const vector<T> &a, T key) {
  ll ng = 0;
  ll ok = (ll)1e9 + 1;

  /* ok と ng のどちらが大きいかわからないことを考慮 */
  while (abs(ok - ng) > 1) {
    ll mid = (ok + ng) / 2;

    if (isOK(a, mid, key)) {
      ok = mid;
    } else {
      ng = mid;
    }
  }
  return ok;
}

void solve(long long N, long long K, std::vector<long long> A) {
  cout << binary_search<ll>(A, K) << endl;
}

int main() {
  long long N;
  scanf("%lld", &N);
  long long K;
  scanf("%lld", &K);
  std::vector<long long> A(N);
  for (int i = 0; i < N; i++) {
    scanf("%lld", &A[i]);
  }
  solve(N, K, std::move(A));
  return 0;
}