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;
long long n, k;
vector<vector<long long>> dp;
vector<pair<long long, long long>> numbers;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  if (n < k) {
    cout << 0 << endl;
    return 0;
  }
  numbers.reserve(n);
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    long long a = 0;
    long long b = 0;
    while (x % 2 == 0) {
      a++;
      x /= 2;
    }
    while (x % 5 == 0) {
      b++;
      x /= 5;
    }
    numbers.push_back({a, b});
  }
  dp.resize(k + 1, vector<long long>(k * 100 + 1, -1));
  dp[0][0] = 0;
  long long max_j = 0;
  for (long long i = 0; i < n; i++) {
    for (long long l = min(k, i + 1); l >= 1; l--) {
      for (long long j = numbers[i].second; j <= k * 100; j++) {
        if (dp[l - 1][j - numbers[i].second] > -1) {
          dp[l][j] = max(dp[l - 1][j - numbers[i].second] + numbers[i].first,
                         dp[l][j]);
        }
      }
    }
  }
  long long ans = 0;
  for (long long j = 0; j <= k * 100; j++) {
    ans = max(ans, min(dp[k][j], j));
  }
  cout << ans << endl;
}
