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 ll long long int
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define GCD(m,n) __gcd(m,n)
#define LCM(m,n) (m*n)/GCD(m,n)
# define mod 1000000007


int main()
{
    IOS;
    ll i,j,k,t,n,sum=0,prod=1,flag=0;
    ll cnt;
    cin >> n >> k;
    ll a[n];
    ll l=1,h=1000000000,mid;
    for(i=0;i<n;i++)
            cin >> a[i];
    ll ans=-1;
    while(l<h)
    {
        mid=(h+l)/2;
        cnt=0;
        for(i=0;i<n;i++)
             {
                 cnt+=(a[i]+mid-1)/mid;
                 cnt--;
             }
        if(cnt<=k)
            h=mid;
        else
           l=mid+1;
    }
    cout << l;
}