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[2][202][5201];
vector<pair<int, int>> v;
int main() {
  int n, k;
  cin >> n >> k;
  v.resize(n, {0, 0});
  for (auto &x : v) {
    long long a;
    cin >> a;
    while (a % 2 == 0) {
      x.first++;
      a /= 2;
    }
    while (a % 5 == 0) {
      x.second++;
      a /= 5;
    }
  }
  fill(dp[0][0], dp[0][0] + 2 * 202 * 5201, -100000);
  dp[0][0][0] = 0;
  for (int i = 0; i < n; i++) {
    fill(dp[(i + 1) & 1][0], dp[(i + 1) & 1][0] + 202 * 5201, -100000);
    for (int j = 0; j <= k; j++)
      for (int p = 0; p <= 5200 - v[i].second; p++) {
        dp[(i + 1) & 1][j][p] = max(dp[(i + 1) & 1][j][p], dp[i & 1][j][p]);
        dp[(i + 1) & 1][j + 1][p + v[i].second] =
            max(dp[(i + 1) & 1][j + 1][p + v[i].second],
                dp[i & 1][j][p] + v[i].first);
      }
  }
  int res = 0;
  for (int i = 0; i < 5200; i++) res = max(res, min(i, dp[n & 1][k][i]));
  cout << res << '\n';
  cin.ignore(2);
  return 0;
}
