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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast,no-stack-protector")
using namespace std;
function<void(void)> ____ = []() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
};
template <typename T>
void operator<<(vector<T> &__container, T x) {
  __container.push_back(x);
}
template <typename T>
ostream &operator<<(ostream &out, vector<T> &__container) {
  for (T _ : __container) out << _ << ' ';
  return out;
}
const int MAXN = 2e5 + 7;
int f[2][205][6007];
void solve() {
  int n, k;
  scanf("%d", &n);
  scanf("%d", &k);
  vector<pair<int, int> > A(n, {0, 0});
  for (int i = 0; i < n; i++) {
    long long int x;
    scanf("%I64d", &x);
    while (x % 2 == 0) x >>= 1, A[i].first++;
    while (x % 5 == 0) x /= 5, A[i].second++;
  }
  int tag = 0;
  memset(f, 255, sizeof(f));
  f[0][0][0] = 0;
  for (int i = 1; i <= n; i++) {
    tag ^= 1;
    memcpy(f[tag], f[tag ^ 1], sizeof f[tag]);
    for (int j = 0; j <= min(i - 1, k - 1); j++) {
      for (int cnt_5 = 0; cnt_5 <= 6000; cnt_5++) {
        if (f[tag ^ 1][j][cnt_5] == -1) continue;
        ((f[tag][j + 1][cnt_5 + A[i - 1].second]) =
             (f[tag][j + 1][cnt_5 + A[i - 1].second]) >
                     (f[tag ^ 1][j][cnt_5] + A[i - 1].first)
                 ? (f[tag][j + 1][cnt_5 + A[i - 1].second])
                 : (f[tag ^ 1][j][cnt_5] + A[i - 1].first));
      }
    }
  }
  int ret = 0;
  for (int i = 1; i <= k; i++)
    for (int j = 0; j <= 6000; j++)
      if (f[tag][i][j] != -1)
        ((ret) =
             (ret) > (min(f[tag][i][j], j)) ? (ret) : (min(f[tag][i][j], j)));
  cout << ret << "\n";
}
int main() {
  solve();
  return 0;
}
