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 solve();
int main() {
  solve();
  return 0;
}
const int maxn = 100010;
const int inf = 0x3f3f3f3f;
const double eps = 1e-5;
const int mod = 1e9 + 7;
int dp[210][20010];
void solve() {
  int n, k;
  cin >> n >> k;
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    int t = 0, f = 0;
    while (x % 2 == 0) t++, x /= 2;
    while (x % 5 == 0) f++, x /= 5;
    for (int p = k; p >= 1; p--) {
      for (int j = t; j <= 20000; j++) {
        if (dp[p - 1][j - t] != -1)
          dp[p][j] = max(dp[p][j], dp[p - 1][j - t] + f);
      }
    }
  }
  int ans = 0;
  for (int i = 1; i <= 20000; i++) {
    if (dp[k][i] != -1) ans = max(ans, min(i, dp[k][i]));
  }
  cout << ans << endl;
}
