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 = 202;
int a[N][2];
int dp[2][N][5003];
int main() {
  ios_base::sync_with_stdio(false);
  int n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    long long v;
    cin >> v;
    int cnt = 0;
    while (v % 2 == 0) {
      v /= 2;
      cnt++;
    }
    a[i][0] = cnt;
    cnt = 0;
    while (v % 5 == 0) {
      v /= 5;
      cnt++;
    }
    a[i][1] = cnt;
  }
  int ans = 0;
  memset(dp, -1, sizeof dp);
  for (int i = 0; i <= n; i++) dp[0][i][0] = 0;
  for (int z = 1; z <= k; z++) {
    int s = z % 2;
    for (int i = 1; i <= n; i++) {
      int t = a[i][0], f = a[i][1];
      for (int j = 0; j <= 5000; j++) {
        dp[s][i][j] = dp[s][i - 1][j];
        if (f <= j) {
          if (dp[!s][i - 1][j - f] != -1)
            dp[s][i][j] = max(dp[!s][i - 1][j - f] + t, dp[s][i][j]);
        }
        ans = max(ans, min(j, dp[s][i][j]));
      }
    }
  }
  cout << ans;
  return 0;
}
