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>
const long long MOD = 1e9 + 7;
using namespace std;
int lim = 0;
long long n, k;
long long v[210][2] = {}, dp[210][30 * 201] = {};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) {
    long long a;
    cin >> a;
    while (a > 0 && a % 2 == 0) {
      v[i][0]++;
      a /= 2;
    }
    while (a > 0 && a % 5 == 0) {
      v[i][1]++;
      a /= 5;
    }
    lim += v[i][1];
  }
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= lim; j++) {
      dp[i][j] = -1000000000000000000;
    }
  }
  long long ans = 0;
  dp[0][0] = 0;
  for (long long i = 1; i <= n; i++) {
    for (long long j = min(i, k); j >= 1; j--) {
      for (long long p = v[i][1]; p <= lim; p++) {
        dp[j][p] = max(dp[j][p], dp[j - 1][p - v[i][1]] + v[i][0]);
      }
    }
  }
  for (long long i = 1; i <= lim; i++) {
    ans = max(ans, min(i, dp[k][i]));
  }
  cout << ans << '\n';
  return 0;
}
