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 a[205], b[205], n, k, ans, dp[205][205 * 40 + 33];
int main(void) {
  long long x;
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &x);
    int u = 0, v = 0;
    while (x % 5 == 0) u++, x /= 5;
    while (x % 2 == 0) v++, x /= 2;
    for (int t = k - 1; t >= 0; t--)
      for (int j = n * 40; j >= 0; j--)
        if (dp[t][j] != -1)
          dp[t + 1][j + u] = max(dp[t + 1][j + u], dp[t][j] + v);
  }
  for (int i = 0; i <= n * 40; i++) ans = max(ans, min(dp[k][i], i));
  printf("%d\n", ans);
  return 0;
}
