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 long long MX = 26 * 205, INF = 1e6;
long long dp[205][MX];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n = 0, k = 0;
  cin >> n >> k;
  vector<pair<long long, long long> > ft;
  long long a = 0;
  for (long long i = 0; i < n; i++) {
    cin >> a;
    long long curt = 0, curf = 0;
    for (long long j = 2; j <= 5; j++) {
      while (a % j == 0) {
        a /= j;
        curt += (j == 2);
        curf += (j == 5);
      }
    }
    ft.push_back(make_pair(curt, curf));
  }
  for (long long i = 0; i <= k; i++)
    for (long long j = 1; j < MX; j++) dp[i][j] = -INF;
  for (long long i = 0; i < n; i++) {
    long long x = ft[i].second, y = ft[i].first;
    for (long long j = min(k - 1, i); j >= 0; j--)
      for (long long l = 0; l + x < MX; l++)
        dp[j + 1][l + x] = max(dp[j + 1][l + x], dp[j][l] + y);
  }
  long long ans = 0;
  for (long long i = 0; i < MX; i++) ans = max(ans, min(dp[k][i], i));
  cout << ans;
  return 0;
}
