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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#pragma GCC optimize("unroll-loops,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
using namespace __gnu_pbds;
using namespace std;
/*
int dx[]={-1,1,0,0};
int dy[]={0,0,-1,1};
*/
#define ull unsigned long long int
#define ll long long int
#define MAX 1000000007
//1e9+7
#define pll pair<ll,ll>
#define pii pair<int,int>
#define IO ios_base::sync_with_stdio(false); cin.tie(NULL);
#define FO ifstream fin("a.in");ofstream fout("a.out");
#define TEST int testcase;cin>>testcase;while(testcase--)
#define all(x) (x).begin(),(x).end()
#define mset0(x) memset((x), 0, sizeof((x)));
#define mset1(x) memset((x), -1, sizeof((x)));
#define endl "\n"
#define plll pair<ll,pll>
typedef tree<ll,null_type,less<ll>,rb_tree_tag, tree_order_statistics_node_update> indexed_set;
FO
ll n,k;
int arr[200001];
bool ok(int m){
    int cnt=0;
    for(int i=0;i<n;i++){
        //if(arr[i]<=m) continue;
        if(arr[i]%m==0) cnt+=arr[i]/m-1;
        else cnt+=arr[i]/m;
    }

    return(cnt<=k);
}
int bsearch(int l,int r){
    if(l==r) return(l);
    if(l+1==r){
        if(ok(l)) return(l);
        return(r);
    }
    int mid=r-(r-l)/2;
    if(ok(mid)) return(bsearch(l,mid));
    else return(bsearch(mid+1,r));
}
void solve(){
    cin>>n>>k;
    int l=1,r=0;
    for(int i=0;i<n;i++) {cin>>arr[i];r=max(r,arr[i]);}
    int ans=bsearch(l,r);
    cout<<ans;
    return;
}
int main(){
    IO
    //TEST
    solve();
    return(0);
}