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;
long long ara[N + 5];
int two[N + 5], five[N + 5];
int n, k;
int dp[2][N + 2][N * 26];
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> ara[i];
  for (int i = 1; i <= n; i++) {
    while (ara[i] % 2 == 0) {
      ara[i] /= 2;
      two[i]++;
    }
    while (ara[i] % 5 == 0) {
      ara[i] /= 5;
      five[i]++;
    }
  }
  for (int i = 0; i <= k; i++) {
    for (int j = 0; j < N * 26; j++) {
      if (i == 0 and j == 0)
        dp[(n + 1) & 1][i][j] = 0;
      else
        dp[(n + 1) & 1][i][j] = -1e5;
    }
  }
  for (int i = n; i >= 1; i--) {
    for (int j = 0; j <= k; j++) {
      for (int l = 0; l < N * 26; l++) {
        dp[i & 1][j][l] = dp[(i + 1) & 1][j][l];
        if (j > 0 and l - five[i] >= 0)
          dp[i & 1][j][l] = max(dp[i & 1][j][l],
                                two[i] + dp[(i + 1) & 1][j - 1][l - five[i]]);
      }
    }
  }
  int ans = 0;
  for (int i = 1; i < N * 26; i++) ans = max(ans, min(i, dp[1][k][i]));
  cout << ans;
}
