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;
long long a[210], b[210];
long long dp[210][10010];
long long n, m, x, ans;
void djj_lxy() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  memset(dp, -1, sizeof dp);
  dp[0][0] = 0;
  cin >> n >> m;
  for (register long long i = 1; i <= n; i++) {
    cin >> x;
    while (x % 2 == 0) x /= 2, a[i]++;
    while (x % 5 == 0) x /= 5, b[i]++;
  }
  for (register long long i = 1; i <= n; i++)
    for (register long long j = m; j >= 0; j--)
      for (register long long k = 10000; k >= 0; k--)
        if (j >= 1 && k >= b[i] && dp[j - 1][k - b[i]] != -1)
          dp[j][k] = max(dp[j][k], dp[j - 1][k - b[i]] + a[i]);
  for (register long long i = 0; i <= 10000; i++)
    ans = max(ans, min(i, dp[m][i]));
  cout << ans;
}
signed main() { djj_lxy(); }
