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 read() {
  long long a = 0, b = getchar(), c = 1;
  while (!isdigit(b)) c = b == '-' ? -1 : 1, b = getchar();
  while (isdigit(b)) a = a * 10 + b - '0', b = getchar();
  return a * c;
}
long long n, k, ans, a[205], dp[205][20005];
int main() {
  n = read(), k = read();
  memset(dp, -1, sizeof(dp)), dp[0][0] = 0;
  for (int i = 0; i < n; i++) {
    long long x = read(), y = 0, z = 0;
    while (x % 2 == 0) y++, x /= 2;
    while (x % 5 == 0) z++, x /= 5;
    for (int j = k - 1; j >= 0; j--)
      for (int p = 0; p < 10005; p++)
        if (~dp[j][p]) dp[j + 1][p + z] = max(dp[j + 1][p + z], dp[j][p] + y);
  }
  for (int i = 0; i < 10005; i++) ans = max(ans, min((long long)i, dp[k][i]));
  printf("%lld", ans);
  return 0;
}
