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;
const int mxn = 201, mxb = 60;
int dp[mxn * mxb][mxn];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  for (int l = 1, a, b; l <= n; ++l) {
    long long x;
    scanf("%lld", &x);
    for (a = 0; x % 2 == 0; x /= 2, ++a)
      ;
    for (b = 0; x % 5 == 0; x /= 5, ++b)
      ;
    for (int j = k - 1; j >= 0; --j) {
      for (int i = l * 60; i >= a; --i)
        if (dp[i - a][j] != -1)
          dp[i][j + 1] = max(dp[i][j + 1], b + dp[i - a][j]);
    }
  }
  int ret = 0;
  for (int i = 0; i <= 60 * n; ++i) ret = max(ret, min(i, dp[i][k]));
  printf("%d\n", ret);
}
