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 dp[205][15010];
long long a[205];
int s2(long long x) {
  int num = 0;
  while (x % 2 == 0) {
    num++;
    x /= 2;
  }
  return num;
}
int s5(long long x) {
  int num = 0;
  while (x % 5 == 0) {
    num++;
    x /= 5;
  }
  return num;
}
int main() {
  int n, k, ans = 0;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
  memset(dp, -0x3f3f3f3f, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    int sum2 = s2(a[i]), sum5 = s5(a[i]);
    for (int j = k; j >= 1; j--)
      for (int l = sum2; l <= 15000; l++)
        dp[j][l] = max(dp[j][l], dp[j - 1][l - sum2] + sum5);
  }
  for (int i = 0; i <= 15000; i++) ans = max(ans, min(dp[k][i], i));
  printf("%d\n", ans);
  return 0;
}
