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;
#define dbg(var) cout<<#var<<"="<<var<<" "
#define nl cout<<"\n"
#define fr(i,n) for(int i=0;i<n;i++)
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define vi vector<int>
#define vvi vector<vi>
#define pb push_back
#define fa(v) for(auto &i:v)
#define all(v) v.begin(),v.end()
#define sz(v) (int)(v.size())
#define int long long
//////////////   debugger class  starts from here///// 
void __print(int x) {cout << x;}
void __print(long x) {cout << x;}
void __print(unsigned x) {cout << x;}
void __print(unsigned long x) {cout << x;}
void __print(unsigned long long x) {cout << x;}
void __print(float x) {cout << x;}
void __print(double x) {cout << x;}
void __print(long double x) {cout << x;}
void __print(char x) {cout << '\'' << x << '\'';}
void __print(const char *x) {cout << '\"' << x << '\"';}
void __print(const string &x) {cout << '\"' << x << '\"';}
void __print(bool x) {cout << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cout << '{'; __print(x.first); cout << ','; __print(x.second); cout << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cout << '{'; for (auto &i: x) cout << (f++ ? "," : ""), __print(i); cout << "}";}
void _print() {cout << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cout << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cout << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif

///////  debugger class ends here /////////////
void solve(){
  int n,k; cin >> n >> k;
  vi v(n); fr(i,n) cin >> v[i];
  int l = 1, r = *max_element(all(v));
  int ans = -1;
  while(l<=r){
  	int len = (l + r) / 2;
  	int cnt = 0;
  	for(int x: v){
  		cnt += (x + len -1 ) / len- 1;
  	}
  	// debug(cnt,len);
  	if(cnt <= k) {
  		ans = len;
  		r = len - 1;
  	}
  	else l = len + 1;
  }
  cout << ans;

}
int32_t main()
{
   ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
   // int tst; cin >> tst; while(tst--)
   {
   	solve();
   }
}