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;
pair<int, int> a[204];
int n, K, f[204][5650], tf[204][5650];
long long x;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> K;
  for (int i = 1; i <= n; ++i) {
    cin >> x;
    while (x > 1 && x % 2 == 0) ++a[i].first, x /= 2;
    while (x > 1 && x % 5 == 0) ++a[i].second, x /= 5;
  }
  for (int i = 0; i <= n; ++i)
    for (int j = 0; j <= 28 * n; ++j) f[i][j] = -1;
  f[0][0] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= K; ++j)
      for (int k = 0; k <= 28 * n; ++k) tf[j][k] = -1;
    for (int j = 1; j <= K; ++j)
      for (int k = a[i].second; k <= 28 * n; ++k)
        if (f[j - 1][k - a[i].second] > -1)
          tf[j][k] = max(tf[j][k], f[j - 1][k - a[i].second] + a[i].first);
    for (int j = 0; j <= K; ++j)
      for (int k = 0; k <= 28 * n; ++k) f[j][k] = max(f[j][k], tf[j][k]);
  }
  int res = 0;
  for (int i = 0; i <= 28 * n; ++i) res = max(res, min(i, f[K][i]));
  cout << res;
}
