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 main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, k;
  vector<pair<int, int> > v;
  cin >> n >> k;
  for (int i = 1; i < n + 1; i++) {
    long long int x;
    cin >> x;
    int cnt = 0, cnt1 = 0;
    while (x % 2 == 0) {
      cnt++;
      x /= 2;
    }
    while (x % 5 == 0) {
      cnt1++;
      x /= 5;
    }
    v.push_back({cnt1, cnt});
  }
  int dp[2][k + 2][9005];
  memset(dp, -1, sizeof(dp));
  dp[0][0][0] = 0, dp[1][0][0] = 0;
  for (int i = 1; i < n + 1; i++)
    for (int j = 1; j < k + 1; j++)
      for (int pwr = 0; pwr < 9001; pwr++) {
        if (pwr >= v[i - 1].first and
            dp[(i - 1) % 2][j - 1][pwr - v[i - 1].first] != -1)
          dp[i % 2][j][pwr] = max(
              dp[i % 2][j][pwr],
              dp[(i - 1) % 2][j - 1][pwr - v[i - 1].first] + v[i - 1].second);
        dp[i % 2][j][pwr] = max(dp[i % 2][j][pwr], dp[(i - 1) % 2][j][pwr]);
      }
  int ans = 0;
  for (int pwr = 0; pwr < 9000; pwr++)
    ans = max(ans, min(pwr, dp[n % 2][k][pwr]));
  cout << ans << endl;
}
