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, kol, koll, z, ans, dp[202][6010];
vector<pair<long long, long long> > a;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> z;
    kol = 0;
    while (z % 2 == 0) {
      z /= 2;
      kol++;
    }
    koll = 0;
    while (z % 5 == 0) {
      z /= 5;
      koll++;
    }
    a.push_back({kol, koll});
  }
  for (int j = 0; j <= k; j++)
    for (int l = 0; l <= 6000; l++) dp[j][l] = -1e8;
  dp[0][0] = 0;
  for (int i = 0; i < n; i++)
    for (int j = k - 1; j >= 0; j--)
      for (int l = 0; l <= 6000; l++)
        if (dp[j][l] >= 0)
          dp[j + 1][l + a[i].second] =
              max(dp[j + 1][l + a[i].second], dp[j][l] + a[i].first);
  for (long long i = 0; i <= 6000; i++) ans = max(ans, min(i, dp[k][i]));
  cout << ans;
  return 0;
}
