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;
int n, k;
int a5[220], a2[220];
int dp[220][6000];
int main() {
  while (cin >> n >> k) {
    memset(a5, 0, sizeof(a5));
    memset(a2, 0, sizeof(a2));
    int sum = 0;
    for (int i = 1; i <= n; i++) {
      long long x;
      cin >> x;
      while (x % 2 == 0) {
        x /= 2;
        a2[i]++;
      }
      while (x % 5 == 0) {
        x /= 5;
        a5[i]++;
      }
      sum += a5[i];
    }
    memset(dp, -0x3f3f3f3f, sizeof(dp));
    dp[0][0] = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = i; j >= 1; j--) {
        for (int p = a5[i]; p <= sum; p++) {
          dp[j][p] = max(dp[j][p], dp[j - 1][p - a5[i]] + a2[i]);
        }
      }
    }
    int ans = 0;
    for (int i = 1; i <= sum; i++) {
      ans = max(ans, min(dp[k][i], i));
    }
    cout << ans << endl;
  }
}
