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;
using ll = long long;
const int INF = 1e9;
const int MAX = 6000;
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> a(n), b(n);
  for (int i = 0; i < n; i++) {
    ll x;
    cin >> x;
    while (x % 2 == 0) ++a[i], x /= 2;
    while (x % 5 == 0) ++b[i], x /= 5;
  }
  vector<vector<int>> dp(k + 1, vector<int>(MAX + 1, -INF));
  dp[0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = k - 1; j >= 0; j--) {
      for (int l = 0; l + b[i] <= MAX; l++) {
        dp[j + 1][l + b[i]] = max(dp[j + 1][l + b[i]], dp[j][l] + a[i]);
      }
    }
  }
  int res = 0;
  for (int i = 0; i <= MAX; i++) {
    res = max(res, min(i, dp[k][i]));
  }
  cout << res << endl;
  return 0;
}
