Let's call the roundness of the number the number of zeros to which it ends.

You have an array of n numbers. You need to choose a subset of exactly k numbers so that the roundness of the product of the selected numbers will be maximum possible.

Input

The first line contains two integer numbers n and k (1 ≤ n ≤ 200, 1 ≤ k ≤ n).

The second line contains n space-separated integer numbers a1, a2, ..., an (1 ≤ ai ≤ 1018).

Output

Print maximal roundness of product of the chosen subset of length k.

Examples

Input

3 2
50 4 20


Output

3


Input

5 3
15 16 3 25 9


Output

3


Input

3 3
9 77 13


Output

0

Note

In the first example there are 3 subsets of 2 numbers. [50, 4] has product 200 with roundness 2, [4, 20] — product 80, roundness 1, [50, 20] — product 1000, roundness 3.

In the second example subset [15, 16, 25] has product 6000, roundness 3.

In the third example all subsets has product with roundness 0.
#include <bits/stdc++.h>
using namespace std;
template <typename T>
ostream &operator<<(ostream &cout, vector<T> &a) {
  for (size_t i = 0; i < a.size(); ++i) cout << a[i] << " ";
  return cout;
}
template <typename T>
ostream &operator<<(ostream &cout, vector<vector<T> > &a) {
  for (size_t i = 0; i < a.size(); ++i) cout << a[i] << endl;
  return cout;
}
int ans[2][201][25 * 200 + 1];
int divis[2][200];
int main() {
  ios_base::sync_with_stdio(0);
  int n, k;
  cin >> n >> k;
  long long a;
  for (int i = 0; i < n; ++i) {
    cin >> a;
    while (a % 2 == 0) {
      divis[0][i]++;
      a /= 2;
    }
    while (a % 5 == 0) {
      divis[1][i]++;
      a /= 5;
    }
  }
  for (int i = 0; i <= k; ++i) {
    for (int j = 0; j <= 25 * 200; ++j) {
      ans[0][i][j] = INT_MIN;
      ans[1][i][j] = INT_MIN;
    }
  }
  ans[0][0][0] = 0;
  ans[0][1][divis[1][0]] = divis[0][0];
  for (int i = 1; i < n; ++i) {
    for (int j = 0; j <= k; ++j) {
      for (int l = 0; l <= 25 * 200; ++l) {
        ans[i % 2][j][l] = max(ans[i % 2][j][l], ans[(i + 1) % 2][j][l]);
        if (j + 1 <= k && l + divis[1][i] <= 25 * 200)
          ans[i % 2][j + 1][l + divis[1][i]] =
              max(ans[(i + 1) % 2][j + 1][l + divis[1][i]],
                  ans[(i + 1) % 2][j][l] + divis[0][i]);
      }
    }
  }
  int res = 0;
  for (int i = 0; i <= 25 * 200; ++i) {
    res = max(res, min(ans[(n - 1) % 2][k][i], i));
  }
  cout << res << endl;
  return 0;
}
