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;
const int N = 210;
int n, k, m;
long long a[N];
int b[N];
int f[N][6010];
int ans = 0;
int c[N], d[N];
int main() {
  scanf("%d%d", &n, &m);
  for (int i(1); i <= (n); ++i) scanf("%lld", a + i);
  for (int i(1); i <= (n); ++i) {
    for (; a[i] % 2 == 0; a[i] /= 2) ++d[i];
    for (; a[i] % 5 == 0; a[i] /= 5) ++c[i];
  }
  for (int j(0); j <= (m); ++j)
    for (int k(0); k <= (5200); ++k) f[j][k] = -(1 << 30);
  f[0][0] = 0;
  for (int i(1); i <= (n); ++i) {
    for (int j(m); j >= (1); --j) {
      for (int k(c[i]); k <= (5200); ++k)
        f[j][k] = max(f[j][k], f[j - 1][k - c[i]] + d[i]);
    }
  }
  for (int j(0); j <= (5200); ++j) ans = max(ans, min(f[m][j], j));
  printf("%d\n", ans);
  return 0;
}
