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 int INF = 2e9 + 5;
int main() {
  cin.tie(0), cin.sync_with_stdio(0);
  int n, k;
  cin >> n >> k;
  vector<long long> v(n);
  for (long long &i : v) cin >> i;
  vector<int> two(n), five(n);
  for (int i = 0; i < n; i++) {
    long long x = v[i];
    while (x % 2 == 0) {
      two[i]++;
      x /= 2;
    }
    while (x % 5 == 0) {
      five[i]++;
      x /= 5;
    }
  }
  const int MAXLOG = 26;
  int maxFives = MAXLOG * n;
  vector<vector<int> > prev(k + 1, vector<int>(maxFives, -1));
  int ans = 0;
  prev[0][0] = 0;
  for (int i = n - 1; i >= 0; i--) {
    vector<vector<int> > curr = prev;
    for (int j = 0; j <= min(k, n - i); j++) {
      for (int f = 0; f < maxFives; f++)
        if (prev[j][f] != -1) curr[j][f] = prev[j][f];
      if (j)
        for (int f = five[i]; f < maxFives; f++) {
          if (prev[j - 1][f - five[i]] != -1) {
            curr[j][f] = max(curr[j][f], two[i] + prev[j - 1][f - five[i]]);
            ans = max(ans, min(f, curr[j][f]));
          }
        }
    }
    swap(prev, curr);
  }
  cout << ans;
}
