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;
long long a[205];
int er[205], wu[205];
int dp[205][6005];
void prin() {
  for (int t = 0; t <= 4; t++) {
    for (int j = 0; j <= k; j++) {
      if (t == 0 && j == k - 1) cout << "(";
      cout << dp[j][t] << " ";
      if (t == 0 && j == k - 1) cout << ") ";
    }
    cout << endl;
  }
}
int main() {
  memset(dp, -1, sizeof(dp));
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    long long t = a[i];
    while (t % 5 == 0) wu[i]++, t /= 5;
    t = a[i];
    while (t % 2 == 0) er[i]++, t /= 2;
  }
  dp[0][0] = 0;
  int ans = -1;
  for (int i = 0; i < n; i++) {
    for (int t = 6000; t >= wu[i]; t--) {
      for (int j = k; j >= 1; j--) {
        if (dp[j - 1][t - wu[i]] != -1)
          dp[j][t] = max(dp[j][t], dp[j - 1][t - wu[i]] + er[i]);
        if (j == k && dp[j][t] != -1) ans = max(ans, min(t, dp[j][t]));
      }
    }
  }
  cout << ans;
  return 0;
}
