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[222];
int two[222], five[222];
int dp[221][7221], ans;
void init(int id) {
  long long cur = a[id];
  while (cur % 2 == 0) {
    cur /= 2;
    ++two[id];
  }
  cur = a[id];
  while (cur % 5 == 0) {
    cur /= 5;
    ++five[id];
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i <= k; ++i) {
    for (int j = 0; j <= 36 * n; ++j) dp[i][j] = -1e9;
  }
  dp[0][0] = 0;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    init(i);
    for (int j = k - 1; j >= 0; --j) {
      for (int cnt = 36 * n; cnt >= 0; --cnt) {
        if (dp[j][cnt] != -1e9) {
          dp[j + 1][cnt + five[i]] =
              max(dp[j + 1][cnt + five[i]], dp[j][cnt] + two[i]);
        }
      }
    }
  }
  for (int cnt = 0; cnt <= 36 * n; ++cnt) {
    ans = max(ans, min(cnt, dp[k][cnt]));
  }
  cout << ans;
  return 0;
}
