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 maxn = 210;
const long long maxm = 210;
const long long logsize = 60;
const long long nlogsize = maxn * logsize;
long long n, m, ans;
long long a[maxn], cnt2[maxn], cnt5[maxn];
long long f[maxm][nlogsize];
signed main() {
  scanf("%I64d%I64d", &n, &m);
  for (long long i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  for (long long i = 1; i <= n; i++) {
    long long tmp = a[i];
    while (tmp % 2 == 0) cnt2[i]++, tmp /= 2;
    while (tmp % 5 == 0) cnt5[i]++, tmp /= 5;
  }
  memset(f, 0xcf, sizeof(f));
  f[0][0] = 0;
  for (long long i = 1; i <= n; i++)
    for (long long j = m; j >= 1; j--)
      for (long long k = 10000; k >= cnt5[i]; k--)
        f[j][k] = max(f[j][k], f[j - 1][k - cnt5[i]] + cnt2[i]);
  for (long long i = 0; i <= 10000; i++) ans = max(ans, min(i, f[m][i]));
  printf("%I64d", ans);
  return 0;
}
