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 x;
int a[210], b[210], k, n, dp1[210][5100], dp2[210][5100], INF = 100000000;
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> x;
    while (x % 5ll == 0) {
      x /= 5ll;
      a[i]++;
    }
    while (x % 2ll == 0) {
      x /= 2ll;
      b[i]++;
    }
  }
  for (int i = 0; i <= k; ++i)
    for (int j = 0; j < 5100; ++j) dp1[i][j] = -INF;
  dp1[0][0] = 0;
  for (int pos = 1; pos <= n; ++pos) {
    for (int i = 0; i <= k; ++i)
      for (int j = 0; j < 5100; ++j) dp2[i][j] = dp1[i][j];
    for (int i = 1; i <= k; ++i)
      for (int j = a[pos]; j < 5100; ++j) {
        dp1[i][j] = max(dp1[i][j], dp2[i - 1][j - a[pos]] + b[pos]);
        dp1[i][j] = max(dp1[i][j], -INF);
      }
  }
  int ans = 0;
  for (int i = 0; i < 5100; ++i)
    if (dp1[k][i] >= 0) ans = max(ans, min(dp1[k][i], i));
  cout << ans << endl;
  return 0;
}
