#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& a) {
  out << a.first << ' ' << a.second;
  return out;
}
const int INF = 0x3f3f3f3f;
const int64_t LINF = 0x3f3f3f3f3f3f3f3fLL;
const int MAX = 205;
const int LOG5 = 26;
int DP[MAX][MAX * LOG5];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> A(n, 0), B(n, 0);
  for (int i = 0; i < n; i++) {
    int64_t x;
    cin >> x;
    while (!(x % 5)) {
      A[i]++;
      x /= 5;
    }
    while (!(x & 1)) {
      B[i]++;
      x >>= 1;
    }
  }
  memset(DP, -1, sizeof DP);
  DP[0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int l = i; l >= 0; l--) {
      for (int x = MAX * LOG5 - 1; x >= 0; x--) {
        if (DP[l][x] == -1) continue;
        if (x + A[i] >= MAX * LOG5) continue;
        DP[l + 1][x + A[i]] = max(DP[l + 1][x + A[i]], DP[l][x] + B[i]);
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < MAX * LOG5; i++) {
    int x = DP[k][i];
    ans = max(ans, min(x, i));
  }
  cout << ans << '\n';
  return 0;
}
