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;
int dp[205][10000];
int get_prime(long long num, long long k) {
  int rt = 0;
  while ((num % k) == 0) {
    rt++;
    num /= k;
  }
  return rt;
}
int main() {
  memset(dp, -1, sizeof(dp));
  int n, K;
  cin >> n >> K;
  dp[0][0] = 0;
  for (int i = 0; i < n; i++) {
    long long num;
    cin >> num;
    int c5 = get_prime(num, 5), c2 = get_prime(num, 2);
    for (int j = K - 1; j >= 0; j--)
      for (int k = 0; k < 10000 - c5; k++)
        if (dp[j][k] != -1)
          dp[j + 1][k + c5] = max(dp[j + 1][k + c5], dp[j][k] + c2);
  }
  int ans = 0;
  for (int i = 0; i < 10000; i++) {
    ans = max(ans, min(i, dp[K][i]));
  }
  cout << ans << endl;
  return 0;
}
