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 inf = 1e9 + 7;
int a[300], b[300], dp[2][300][5210];
int main() {
  int n, k, ans = 0;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    while (x % 5 == 0) {
      a[i]++;
      x /= 5;
    }
    while (x % 2 == 0) {
      b[i]++;
      x /= 2;
    }
  }
  for (int j = 0; j <= k; j++) {
    for (int l = 0; l <= 5200; l++) dp[0][j][l] = dp[1][j][l] = -inf;
  }
  dp[0][0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= k; j++) {
      for (int l = 0; l <= 5200; l++) {
        dp[(i + 1) % 2][j][l] = max(dp[(i + 1) % 2][j][l], dp[i % 2][j][l]);
        int x = l + a[i];
        dp[(i + 1) % 2][j + 1][x] =
            max(dp[(i + 1) % 2][j + 1][x], b[i] + dp[i % 2][j][l]);
      }
    }
  }
  for (int i = 0; i <= 5200; i++) ans = max(ans, min(i, dp[n % 2][k][i]));
  cout << ans;
  return 0;
}
