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;
int max2[205][6005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  fill_n(max2[0], 205 * 6005, -1);
  max2[0][0] = 0;
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    long long int a;
    cin >> a;
    int p2 = 0, p5 = 0;
    while (a % 2 == 0) {
      a /= 2;
      p2++;
    }
    while (a % 5 == 0) {
      a /= 5;
      p5++;
    }
    for (int pos = k - 1; pos >= 0; pos--) {
      for (int s = p5; s < 6005; s++) {
        if (max2[pos][s - p5] != -1) {
          max2[pos + 1][s] = max(max2[pos + 1][s], max2[pos][s - p5] + p2);
        }
      }
    }
  }
  int best = 0;
  for (int i = 0; i < 6005; i++) {
    best = max(best, min(i, max2[k][i]));
  }
  cout << best << '\n';
  return 0;
}
