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: Wrong Answer)

#include <bits/stdc++.h>
#define REP(i, n) for(int i = 0; i < (n); i++)
#define REP1(i, n) for(int i = 1; i <= (n); i++)
#define REPD(i,a,b) for (int i=(a);i<=(b);i++)
#define ALL(v) (v).begin(), (v).end()
using namespace std;
typedef long long ll;
typedef vector<string> vs;
typedef vector<bool> vb;
typedef vector<vb> vvb;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef set<int> si;
typedef vector<si> vsi;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef pair<int, int> pi;
typedef queue<int> qi;
typedef queue<pi> qpi;
typedef pair<ll, ll> pll;
typedef queue<pll> qpll;
typedef vector<pi> vpi;
typedef vector<pll> vpll;
const int mod = 1000000007;
const int INF = 1001001001;
// 小数点　<< fixed << setprecision(0) <<
// sort降順　sort(ALL(),greater<int>());
// 円周率　M_PI
// 文字判定　isupper islower
// 文字変換　
// 順列　do {} while(next_permutation(ALL(X)));
// 最大値　LLONG_MAX
// a内でx以上　auto iter = lower_bound(ALL(a), x);
// a内でxより大きい　auto iter = upper_bound(ALL(a), x);
int dx[8] = {1, 0, -1, 0, 1, 1, -1, -1};
int dy[8] = {0, 1, 0, -1, 1, -1, 1, -1};
struct edge {
    int from; //出発点
    int to;   //到達点
    int cost; //移動コスト
};
typedef struct edge se;
typedef vector<edge> ve;
unsigned Euclidean_gcd(unsigned a, unsigned b) {
  if(a < b) return Euclidean_gcd(b, a);
  unsigned r;
  while ((r=a%b)) {
    a = b;
    b = r;
  }
  return b;
}

int digit_sum(int n) {
  int sum=n%10;
  while(n!=0) {
    n/=10;
    sum+=n%10;
  }
  return sum;
}

vpll PrimeFactorization(ll n) {
  vpll res;
  for (ll i=2; i*i<=n; i++) {
    if(n%i!=0) continue;
    ll ex=0;
    while(n%i==0) {
      ex++;
      n/=i;
    }
    res.push_back(pll(i,ex));
  }
  if(n!=1) res.push_back(pll(n,1));
  return res;
}


ll n,k;
vll a(200001);

int count(double x) {
  int cnt=0;
  REP(i,n) {
    cnt+=ceil(a[i]/x)-1;
  }
  return cnt;
}

int main() {
  cin >> n >> k;
  REP(i,n) cin >> a[i];
  sort(ALL(a),greater<int>());
  double l=0,r=a[0]*2;
  int temp=k+1;
  while(l+1<r) {
    double c=(l+r)/2.0;
    temp=count(c);
    if(temp>k) l=c;
    else r=c; 
  }
  ll ans = ceil(r);
  cout << ans << endl;

  return 0;
}