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 n, k;
long long A[205];
long long C2[205];
long long C5[205];
long long dp[205][10005];
long long solve = 0, c5 = 0, c2 = 0;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
    long long v2 = 0, v5 = 0;
    long long x = A[i];
    while (x % 2ll == 0) x /= 2ll, v2++;
    x = A[i];
    while (x % 5ll == 0) x /= 5ll, v5++;
    C2[i] = v2, C5[i] = v5;
    c5 += v5;
    c2 += v2;
  }
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  for (long long i = 1; i <= n; i++)
    for (long long l = k; l >= 1; l--)
      for (long long j = c5; j >= C5[i]; j--) {
        if (dp[l - 1][j - C5[i]] == -1) continue;
        dp[l][j] = max(dp[l][j], dp[l - 1][j - C5[i]] + C2[i]);
        solve = max(solve, min(j, dp[l][j]));
      }
  cout << solve << '\n';
  return 0;
}
