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;
long long a[205], b[205];
long long f(long long x, long long p) {
  long long res = 0;
  while (x % p == 0) x /= p, res++;
  return res;
}
long long dp[205][20205];
signed main() {
  long long x, n, k, ma = 0, mb = 0;
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) cin >> x, a[i] = f(x, 5), b[i] = f(x, 2);
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  ma = 10005;
  for (long long i = 1; i <= n; i++)
    for (long long j = k; j >= 1; j--)
      for (long long o = ma; o >= a[i]; o--)
        if (dp[j - 1][o - a[i]] != -1)
          dp[j][o] = max(dp[j][o], dp[j - 1][o - a[i]] + b[i]);
  long long ans = 0;
  for (long long i = 0; i <= ma; i++) ans = max(ans, min(i, dp[k][i]));
  cout << ans;
  return 0;
}
