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;
const int N = 200 + 5, LG = 40, inf = 1e9;
pair<int, int> dp[2][N][LG * N];
int n, m, a[N], b[N];
pair<int, int> mx(pair<int, int> x, pair<int, int> y) {
  if (x.first == y.first) return ((x.second > y.second) ? x : y);
  return ((x.first > y.first) ? x : y);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    while (x % 2 == 0) a[i]++, x /= 2;
    while (x % 5 == 0) b[i]++, x /= 5;
  }
  for (int i = 0; i <= 1; i++)
    for (int j = 0; j <= m; j++)
      for (int k = 0; k < N * LG; k++) dp[i][j][k] = {-inf, -inf};
  dp[0][0][0] = {0, 0};
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= min(i + 1, m); j++) {
      for (int k = 0; k <= (i + 1) * LG; k++) {
        dp[(i + 1) & 1][j][k] = dp[i & 1][j][k];
        if (j && k >= b[i]) {
          int mn = min(k, dp[i & 1][j - 1][k - b[i]].second + a[i]);
          dp[(i + 1) & 1][j][k] =
              mx(dp[(i + 1) & 1][j][k],
                 make_pair(mn, dp[i & 1][j - 1][k - b[i]].second + a[i]));
        }
      }
    }
  }
  pair<int, int> ans = {0, 0};
  for (int i = 0; i <= n * LG; i++) ans = mx(ans, dp[n & 1][m][i]);
  cout << ans.first << "\n";
}
