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 n, k, t[205], ans, f[205], dp[205][5005];
bool vis[205][5005];
long long a[205];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    while (a[i] % 2 == 0) {
      a[i] /= 2;
      ++t[i];
    }
    while (a[i] % 5 == 0) {
      a[i] /= 5;
      ++f[i];
    }
  }
  vis[0][0] = true;
  for (int i = 1; i <= n; i++)
    for (int h = min(k, i); h >= 1; h--)
      for (int j = 25 * h; j >= 0; j--)
        if (j - f[i] >= 0 && vis[h - 1][j - f[i]]) {
          dp[h][j] = max(dp[h][j], dp[h - 1][j - f[i]] + t[i]);
          vis[h][j] = true;
        }
  for (int i = 0; i < 5150; i++)
    if (vis[k][i]) ans = max(ans, min(i, dp[k][i]));
  printf("%d", ans);
  return 0;
}
