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 long long INF = 0x7ffffff;
int x[220], y[220];
long long a, ans, f[220][6000];
int main() {
  int n, m, sum = 0;
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a);
    long long now = a;
    while (now % 5 == 0) {
      x[i]++;
      now /= 5;
    }
    sum += x[i];
    now = a;
    while (now % 2 == 0) {
      y[i]++;
      now >>= 1;
    }
  }
  for (int k = 0; k <= m; k++)
    for (int j = 0; j < 5200; j++) f[k][j] = -INF;
  f[0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int k = m; k > 0; k--)
      for (int j = x[i]; j <= sum; j++)
        f[k][j] = max(f[k][j], f[k - 1][j - x[i]] + y[i]);
  }
  for (int j = 0; j <= sum; j++) {
    long long now = f[m][j];
    if (j < now) now = j;
    ans = max(now, ans);
  }
  printf("%lld\n", ans);
  return 0;
}
