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 N = 205;
int c2[N], c5[N];
long long a[N];
int n, k;
long long ans = 0, dp[N][10005];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &a[i]);
    while (a[i] % 2 == 0 && a[i]) {
      ++c2[i];
      a[i] /= 2;
    }
    while (a[i] % 5 == 0 && a[i]) {
      ++c5[i];
      a[i] /= 5;
    }
  }
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = k; j; --j) {
      for (int l = 10000; l >= c5[i]; --l) {
        if (dp[j - 1][l - c5[i]] != -1)
          dp[j][l] = max(dp[j][l], dp[j - 1][l - c5[i]] + c2[i]);
      }
    }
  }
  for (int i = 0; i <= 10000; ++i) {
    ans = max(ans, min((long long)(i), dp[k][i]));
  }
  printf("%lld\n", ans);
  return 0;
}
