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>
long long get() {
  long long x = 0;
  char s = getchar();
  while (s < '0' || s > '9') s = getchar();
  while (s >= '0' && s <= '9')
    x = (x << 3) + (x << 1) + (s ^ 48), s = getchar();
  return x;
}
long long Max(const long long x, const long long y) {
  if (x > y) return x;
  return y;
}
long long Min(const long long x, const long long y) {
  if (x < y) return x;
  return y;
}
long long a[205], b[205], dp[205][10005], n, m, ans;
signed main() {
  memset(dp, -1, sizeof dp);
  dp[0][0] = 0;
  n = get(), m = get();
  for (long long i = 1, t; i <= n; ++i) {
    t = get();
    while (t % 2 == 0) t >>= 1, ++a[i];
    while (t % 5 == 0) t /= 5, ++b[i];
  }
  for (long long i = 1; i <= n; ++i)
    for (long long j = m; j >= 0; --j)
      for (long long k = 10000; k >= 0; --k)
        if (j && k >= b[i] && ~dp[j - 1][k - b[i]])
          dp[j][k] = Max(dp[j][k], dp[j - 1][k - b[i]] + a[i]);
  for (long long i = 0; i <= 10000; ++i) ans = Max(ans, Min(i, dp[m][i]));
  printf("%lld", ans);
  return 0;
}
