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 double sn = 1e-6;
const int siz = 100005;
int n, k;
pair<int, int> p[205];
long long arr[205];
vector<vector<short>> dp[200];
int solve(int i, int j, int l) {
  if (l < 0 || (j == k && l == 0)) return 0;
  if (i == n || (k - j) * 26 < l || n - i < k - j || j == k) return -2;
  if (dp[i][j][l] != -1) return dp[i][j][l];
  int v1 = solve(i + 1, j, l), v2 = solve(i + 1, j + 1, l - p[i].second);
  if (v2 != -2) v2 += p[i].first;
  return dp[i][j][l] = max(v1, v2);
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &arr[i]);
    long long v1 = arr[i], v2 = arr[i];
    while (v1 % 2 == 0) {
      p[i].first++;
      v1 /= 2;
    }
    while (v2 % 5 == 0) {
      p[i].second++;
      v2 /= 5;
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= min(i, k); j++) {
      dp[i].push_back({});
      for (int l = 0; l <= 26 * (k - j); l++) {
        dp[i][j].push_back(-1);
      }
    }
  }
  int ans = -2;
  for (int i = 0; i <= k * 26; i++) {
    ans = max(ans, min(i, solve(0, 0, i)));
  }
  printf("%d\n", ans);
  return 0;
}
