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 maxn = 207;
int n, K;
long long a[maxn];
int sum5[maxn], sum2[maxn], dp[maxn][40 * maxn];
int main() {
  scanf("%d%d", &n, &K);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    long long tmp = a[i];
    while (tmp % 5 == 0) {
      sum5[i]++;
      tmp /= 5;
    }
    tmp = a[i];
    while (tmp % 2 == 0) {
      sum2[i]++;
      tmp /= 2;
    }
  }
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++)
    for (int j = K; j >= 1; j--)
      for (int k = 6000; k >= 0; k--)
        if (dp[j - 1][k] != -1)
          dp[j][k + sum5[i]] = max(dp[j][k + sum5[i]], dp[j - 1][k] + sum2[i]);
  int ans = 0;
  for (int k = 0; k <= 6000; k++)
    if (dp[K][k] >= 0) ans = max(ans, min(dp[K][k], k));
  printf("%d\n", ans);
  return 0;
}
