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 = 200 + 1, Mod = 1e9 + 7;
int n, m, dp[2][N][N * 30], f[N], t[N], A;
long long a[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    while (a[i] % 2 == 0) {
      t[i]++;
      a[i] /= 2;
    }
    while (a[i] % 5 == 0) {
      f[i]++;
      a[i] /= 5;
    }
  }
  for (int i = 1; i < N * 30; i++)
    for (int j = 0; j < N; j++) dp[1][j][i] = dp[0][j][i] = -Mod;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      for (int k = 0; k < N * 30; k++) {
        dp[i % 2][j][k] = dp[(i - 1) % 2][j][k];
        if (k >= f[i]) {
          dp[i % 2][j][k] =
              max(dp[i % 2][j][k], dp[(i - 1) % 2][j - 1][k - f[i]] + t[i]);
        }
      }
    }
  }
  for (int i = 1; i < N * 30; i++) {
    A = max(A, min(i, dp[n % 2][m][i]));
  }
  cout << A;
  return 0;
}
