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;
long long dp[2][201][5001];
pair<int, int> a[201];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    while (x % 2 == 0) a[i].first++, x /= 2;
    while (x % 5 == 0) a[i].second++, x /= 5;
  }
  for (int i = 0; i <= k; i++)
    for (int l = 0; l <= 5000; l++) dp[1][i][l] = INT_MIN;
  dp[1][0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= k; j++)
      for (int l = 0; l <= 5000; l++) {
        dp[0][j][l] = dp[1][j][l];
        dp[1][j][l] = INT_MIN;
      }
    dp[1][0][0] = 0;
    for (int j = 1; j <= min(k, i + 1); j++) {
      for (int l = 0; l <= 5000; l++) {
        if (a[i].second <= l)
          dp[1][j][l] =
              max(dp[1][j][l], dp[0][j - 1][l - a[i].second] + a[i].first);
        dp[1][j][l] = max(dp[1][j][l], dp[0][j][l]);
      }
    }
  }
  long long ans = 0;
  for (long long i = 0; i <= 5000; i++) ans = max(ans, min(dp[1][k][i], i));
  cout << ans;
  return 0;
}
