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>
struct dta {
  int t2, t5;
  dta(int _t2 = 0, int _t5 = 0) {
    t2 = _t2;
    t5 = _t5;
  }
};
using namespace std;
int n, k, S[201], F[201][201 * 26], Ftam[201][201 * 26];
dta P[201];
dta cal(long long x) {
  int a = 0, b = 0;
  while (!(x % 2)) x /= 2, a++;
  while (!(x % 5)) x /= 5, b++;
  return dta(a, b);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    long long t;
    scanf("%lld", &t);
    P[i] = cal(t);
  }
  int moc = k * 25;
  for (int j = 0; j <= k; j++)
    for (int q = 0; q <= moc; q++) F[j][q] = Ftam[j][q] = -1e9;
  Ftam[0][0] = F[0][0] = 0;
  S[0] = 0;
  for (int i = 1; i <= n; i++) S[i] = S[i - 1] + P[i].t5;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j <= k; j++)
      for (int q = 0; q <= moc; q++) Ftam[j][q] = F[j][q];
    for (int j = 1; j <= k; j++) {
      for (int q = 0; q <= moc; q++) {
        F[j][q] = Ftam[j][q];
        if (q - P[i].t5 >= 0)
          F[j][q] = max(F[j][q], Ftam[j - 1][q - P[i].t5] + P[i].t2);
      }
    }
  }
  int res = 0;
  for (int i = 0; i <= S[n]; i++) res = max(res, min(F[k][i], i));
  printf("%d", res);
  return 0;
}
