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, m, s;
int f[10050][250];
int main() {
  memset(f, 0x80, sizeof(f));
  f[0][0] = 0;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; ++i) {
    long long p;
    int x = 0, y = 0;
    scanf("%I64d", &p);
    for (; !(p % 5); p /= 5) ++x;
    for (; !(p % 2); p /= 2) ++y;
    s += x;
    for (int k = min(i, m); k >= 1; --k)
      for (int j = s; j >= x; --j) f[j][k] = max(f[j][k], f[j - x][k - 1] + y);
  }
  int ans = 0;
  for (int i = 1; i <= s; ++i) ans = max(ans, min(i, f[i][m]));
  printf("%d\n", ans);
  return 0;
}
