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 int64_t INF = 1e18 + 10;
int n, k;
const int N = 4000;
int count_2[200];
int count_5[200];
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int64_t a;
    cin >> a;
    int64_t t = a;
    while (t % 2 == 0) {
      count_2[i]++;
      t /= 2;
    }
    t = a;
    while (t % 5 == 0) {
      count_5[i]++;
      t /= 5;
    }
  }
  static int dp[210][10000];
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  int sum_5 = 0;
  for (int i = 0; i < n; i++) {
    for (int j = k - 1; j >= 0; j--) {
      for (int l = sum_5; l >= 0; l--) {
        if (dp[j][l] != -1) {
          dp[j + 1][l + count_5[i]] =
              max(dp[j + 1][l + count_5[i]], dp[j][l] + count_2[i]);
        }
      }
    }
    sum_5 += count_5[i];
  }
  int ans = 0;
  for (int i = 0; i <= sum_5; i++) {
    ans = max(ans, min(i, dp[k][i]));
  }
  cout << ans << endl;
  return 0;
}
