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: Runtime Error)

/*****************************************************
 *     created by: nayakashutosh9
 * "Winners Never Quit and Quitters Never Win".
*****************************************************/
#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
#pragma GCC optimize("Ofast")
using namespace std;
using namespace __gnu_pbds;
#define int long long int
#define endl '\n'
#define mod 1000000007
#define inf 1e18
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
const int N = 300005;
/**********************************************************************************************/

vector<int> a;
bool check(int n, int cur, int k) {
	int c = 0;
	for (int i = 0; i < n; i++) {
		c += (a[i] / cur);
	}
	return c <= k;
}
void solve() {
	int n, k; cin >> n >> k;
	a.clear(); a.resize(n);
	for (int i = 0; i < n; i++) cin >> a[i];
	int mid, ans, lo = 0, hi = *max_element(a.begin(), a.end());
	ans = hi;
	while (lo <= hi) {
		mid = (lo + hi) / 2;
		if (check(n, mid, k)) {
			ans = mid; hi = mid - 1;
		}
		else lo = mid + 1;
	}
	cout << ans << endl;
}

int32_t main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
	int t = 1, z = 1;
	// cin >> t; cin.ignore();
	while (t--) {
		// cout << "Case #" << (z++) << ": ";
		// clock_t start = clock();
		solve();
		// clock_t end = clock();
		// cout << (end-start) << endl;
		//print execution time in ms
	}

	return 0;
}