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;
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << "\n";
  err(++it, args...);
}
const double PI = acos(-1);
const double eps = (1e-9);
const long long int INF = (1e18);
const int M = (1e9) + 7;
const int N = 203;
const int K = 5250;
int dp[N][K];
int ar[N][K];
void solve() {
  int n, r, i, j, k, a, b;
  scanf("%d%d", &n, &r);
  vector<pair<int, int> > vp;
  long long int x;
  for (i = 0; i < n; ++i) {
    cin >> x;
    a = 0, b = 0;
    while (x % 2 == 0) x /= 2, a++;
    while (x % 5 == 0) x /= 5, b++;
    vp.emplace_back(a, b);
  }
  for (i = 0; i < N; ++i) {
    for (j = 0; j < K; ++j) dp[i][j] = -1, ar[i][j] = -1;
  }
  dp[0][0] = 0, ar[0][0] = 0;
  for (i = 0; i < n; ++i) {
    tie(a, b) = vp[i];
    for (j = 0; j < i + 1; ++j) {
      for (k = 0; k < K - b; ++k) {
        if (dp[j][k] >= 0)
          ar[j + 1][k + b] = max(ar[j + 1][k + b], dp[j][k] + a);
      }
    }
    for (j = 0; j < N; ++j) {
      for (k = 0; k < K; ++k) dp[j][k] = max(dp[j][k], ar[j][k]);
    }
  }
  k = 0;
  for (j = 0; j < K; ++j) {
    if (dp[r][j] >= 0) k = max(k, min(j, dp[r][j]));
  }
  printf("%d\n", k);
}
int main() {
  int t;
  t = 1;
  while (t--) solve();
}
