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 <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <string>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <iomanip>
#include <limits>
using namespace std;
// 素因数分解
typedef pair<long long, long long> P;
vector<P> prime_factorize(long long n)
{
vector<P> res;
for (long long p = 2; p * p <= n; ++p)
{
if (n % p != 0)
continue;
int num = 0;
while (n % p == 0)
{
++num;
n /= p;
}
res.push_back(make_pair(p, num));
}
if (n != 1)
res.push_back(make_pair(n, 1));
return res;
}
// 因数分解
long long gcd(long long x, long long y)
{
  if (y == 0)
  {
    return x;
  }
  else
  {
    return gcd(y, x % y);
  }
}
//素数判定
bool is_prime(long long n){
    if (n <= 1) return false;
    for (long long p = 2; p * p <= n; ++p) {
        if (n % p == 0) return false;
    }
    return true;
}
const int MOD = 1000000007;
typedef long long int ll;
#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
#define rep(i, n) for (int i = 0; i < (n); i++)
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

int main()
{
    int n, k;
    cin >> n >> k;
    vector<int> a(n);
    rep(i,n){
        cin >> a[i];
    }
    int l = 0,r = 1e9;
    while(r-l>1){
        int x = (l + r) / 2;
        auto f = [&](int x) {
        ll now = 0;
        rep(i,n)
        {
            now += (a[i] - 1) / x;
        }
        return now <= k;
        };
        if(f(x)){
            r = x;
        }else{
            l = x;
        }
    }
    cout << r << endl;
    return 0;
}