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;
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;
}
