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)

#include        "algorithm"
#include        "iostream"
#include        "numeric"
#include        "iomanip"
#include        "cstring"
#include        "math.h"
#include        "bitset"
#include        "string"
#include        "vector"
#include        "random"
#include        "chrono"
#include        "ctime"
#include        "queue"
#include        "stack"
#include        "map"
#include        "set"

#include        "ext/pb_ds/assoc_container.hpp" // Common file
#include        "ext/pb_ds/tree_policy.hpp" // Including tree_order_statistics_node_update
#include        "ext/pb_ds/detail/standard_policies.hpp"

using namespace std;
using namespace __gnu_pbds;


#define          f first
#define          lgn 25
#define          endl '\n'
#define          sc second
#define          pb push_back
#define          N (int)5e5+5
#define          PI acos(-1.0)
#define          int long long 
#define          vi vector<int>
#define          mod 1000000007
#define          ld long double
#define          eb emplace_back
#define          mii map<int,int>
#define          vpii vector<pii>
#define          pii pair<int,int>
#define          pq priority_queue
#define          BLOCK (int)sqrt(N)
#define          test(x) while(x--)
#define          all(x) begin(x),end(x)
#define          allr(x) x.rbegin(),x.rend()
#define          fo(i,a,n) for(int i=a;i<n;i++)
#define          rfo(i,n,a) for(int i=n;i>=a;i--)
#define          FAST ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define          time() cerr << "Time : " << (double)clock() / (double)CLOCKS_PER_SEC << "s\n"
#define          bug(...)       __f (#__VA_ARGS__, __VA_ARGS__)

typedef tree< int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update > 
OS ;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

template <typename Arg1>
void __f (const char* name, Arg1&& arg1) { cout << name << " : " << arg1 << endl; }
template <typename Arg1, typename... Args>
void __f (const char* names, Arg1&& arg1, Args&&... args)
{
     const char* comma = strchr (names + 1, ',');
     cout.write (names, comma - names) << " : " << arg1 << " | "; __f (comma + 1, args...);
}

const int inf = 0x3f3f3f3f;
const int INF = 0x3f3f3f3f3f3f3f3f;

int n,m,k,q;
string s;
vi adj[N];
int vis[N],par[N],a[N];

bool check( int mid )
{
    int cn = 0;
    fo(i,0,n)
    {
        if( a[i] < mid ) continue;
        cn += ( ( a[i] + mid - 1) / mid ) - 1;
    }

    return ( cn <= k );
}

void go()
{
    cin >> n >> k;
    int mx = 0;
    fo(i,0,n) cin >> a[i] , mx = max( mx , a[i] );

    int l = 0 , r = 1e11;
    int ans = mx;

    while( l <= r)
    {
        int mid = l + ( r - l )/2;

        if( check( mid ) )
        {
            ans = mid;
            r = mid - 1;
        }
        else l = mid + 1;
    }

    cout << ans << endl;
}

int32_t main()
{
    FAST;     
    int t=1; 
    // cin>>t;
    test(t) go();
}