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::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<pair<int, int> > a(n);
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    int c2 = 0;
    int c5 = 0;
    while (x % 2 == 0) x /= 2, c2 += 1;
    while (x % 5 == 0) x /= 5, c5 += 1;
    a[i] = make_pair(c2, c5);
  }
  const int INF = 1e9;
  vector<vector<int> > dp(n + 1, vector<int>(28 * n, -INF));
  dp[0][0] = 0;
  for (int pos = 0; pos < n; pos++) {
    vector<vector<int> > nxt(n + 1, vector<int>(28 * n, -INF));
    auto [c2, c5] = a[pos];
    for (int taken = 0; taken <= n; taken++) {
      for (int p5 = 0; p5 < 28 * n; p5++) {
        nxt[taken][p5] = max(nxt[taken][p5], dp[taken][p5]);
        if (taken > 0 and p5 >= c5) {
          nxt[taken][p5] = max(nxt[taken][p5], c2 + dp[taken - 1][p5 - c5]);
        }
      }
    }
    dp = nxt;
  }
  int ans = 0;
  for (int x = 1; x < 28 * n; x++) ans = max(ans, min(x, dp[k][x]));
  cout << ans << '\n';
  return (0);
}
