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;
constexpr int inf32 = 0x3f3f3f3f;
constexpr long long inf64 = 0x3f3f3f3f3f3f3f3f;
constexpr int N5 = 8000;
void mymax(int &first, int second) { first = first < second ? second : first; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n, k;
  cin >> n >> k;
  vector<pair<int, int> > a(n);
  for (int i = 0; i < (n); ++i) {
    long long num;
    cin >> num;
    while (num % 2 == 0) a[i].first++, num /= 2;
    while (num % 5 == 0) a[i].second++, num /= 5;
  }
  vector<vector<int> > f(k + 1, vector<int>(N5, -inf32));
  f[0][0] = 0;
  for (int i = 0; i < (n); ++i)
    for (int j = (k)-1; j >= 0; --j)
      for (int v = (N5 - a[i].second) - 1; v >= 0; --v)
        if (f[j][v] != -inf32)
          mymax(f[j + 1][v + a[i].second], f[j][v] + a[i].first);
  int ns = 0;
  for (int v = 0; v < (N5); ++v)
    if (f[k][v] != -inf32) mymax(ns, min(v, f[k][v]));
  cout << ns << '\n';
  return 0;
}
