/* Driver and binary search solution for Leetcode problem "Koko Eating Bananas". */

#include <vector>
#include <iostream>
using namespace std;

class Solution {
public:
    int minEatingSpeed(vector<int>& piles, int h);
};

int Solution::minEatingSpeed(vector<int>& piles, int h)
{
  int start, size, maxpile, mid;
  size_t i;
  long long timesteps, for_pile;

  /* Calculate the maximum pile size. */

  maxpile = piles[0];
  for (i = 0; i < piles.size(); i++) if (piles[i] > maxpile) maxpile = piles[i];

  /* We want our range to start at one and end at maxpile (including maxpile).
     So we set start to 1 and size to maxpile.  Remember size means that start+size is one
     past the last element in our region. */

  start = 1;
  size = maxpile;

  while (size > 1) {

    /* You want to test the highest value in the first half of the values (
       the last value in the blue region of the picture. */

    mid = start + size/2 - 1;
    
    /* Timesteps is the total timesteps if k is set to mid. */

    timesteps = 0;
    for (i = 0; i < piles.size(); i++) {
      for_pile = piles[i] / mid;
      if (piles[i]%mid != 0) for_pile++;
      timesteps += for_pile;
    } 

    // printf("Start: %d.  Size: %d  Mid: %d.  Timesteps: %lld\n", start, size, mid, timesteps);

    /* If timesteps is too big, then you know that the answer
       is in the second half of the range.  You can throw out the
       first half. */

    if (timesteps > h) {
      start += size/2;
      size -= size/2;      

    /* Otherwise, the answer is in the first half of the range, so toss out the second half. */

    } else {
      size = size/2;
    }
  }
  return start;
}

int main()
{
  vector <int> p;
  int h;
  Solution s;

  while (cin >> h) p.push_back(h);   /* Read the piles and h and put them into p */
  p.pop_back();                      /* Remove h from p. */
  cout << s.minEatingSpeed(p, h) << endl;
  return 0;
}