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 readint() {
  char c;
  while (c = getchar(), (c < '0' || c > '9') && c != '-')
    ;
  bool flag = (c == '-');
  if (flag) c = getchar();
  int x = 0;
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - 48;
    c = getchar();
  }
  return flag ? -x : x;
}
inline string tos(int x) {
  string s;
  while (x) s = (char)(x % 10 + '0') + s, x /= 10;
  return s;
}
int dp[2][205][5005];
long long a[205], n, k;
int main() {
  ios_base::sync_with_stdio(false);
  int i, j, l, cnt5, cnt2, ans = 0;
  cin >> n >> k;
  for (i = 1; i <= n; i++) cin >> a[i];
  memset(dp, -1, sizeof(dp));
  dp[0][0][0] = 0;
  for (i = 1; i <= n; i++) {
    cnt2 = cnt5 = 0;
    while (a[i] % 2 == 0) cnt2++, a[i] /= 2;
    while (a[i] % 5 == 0) cnt5++, a[i] /= 5;
    for (j = 0; j <= k; j++)
      for (l = 0; l <= 5000; l++) {
        dp[i & 1][j][l] = dp[(i - 1) & 1][j][l];
        if (j && l >= cnt5 && ~dp[(i - 1) & 1][j - 1][l - cnt5]) {
          dp[i & 1][j][l] =
              max(dp[i & 1][j][l], dp[(i - 1) & 1][j - 1][l - cnt5] + cnt2);
        }
      }
  }
  for (i = 0; i <= 5000; i++) {
    ans = max(ans, min(i, dp[n & 1][k][i]));
  }
  cout << ans << endl;
  return 0;
}
