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 maxm = 205 * 64;
int dp[205][maxm];
int main() {
  int n, k;
  while (~scanf("%d%d", &n, &k)) {
    for (int i = 0; i <= k; i++) {
      for (int j = 0; j <= maxm; j++) dp[i][j] = -0x3f3f3f3f;
      ;
    }
    dp[0][0] = 0;
    for (int i = 0; i < n; i++) {
      long long x, tmp;
      int n2 = 0, n5 = 0;
      scanf("%I64d", &x);
      tmp = x;
      while (x % 2 == 0) {
        n2++;
        x /= 2;
      }
      while (tmp % 5 == 0) {
        n5++;
        tmp /= 5;
      }
      for (int i = k; i >= 1; i--) {
        for (int j = n2; j < maxm; j++)
          dp[i][j] = max(dp[i][j], dp[i - 1][j - n2] + n5);
      }
    }
    int ans = 0;
    for (int i = 1; i <= maxm; i++) {
      ans = max(ans, min(i, dp[k][i]));
    }
    printf("%d\n", ans);
  }
}
