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;
const int mod = 998244353;
const long long INF = -9999999999999;
const int maxn = 1000006;
const int maxm = 1000006;
const int ox3f = 1061109567;
const int fox3f = -2139062144;
const long long ox3fll = 4557430888798830399;
int n, m, T;
int dp[202][5200];
long long c[202];
int a[202], b[202];
void solve() {
  cin >> n >> m;
  int sum1 = 0;
  dp[0][0] = 1;
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
    while (c[i] > 0 && c[i] % 2 == 0) {
      a[i]++;
      c[i] >>= 1;
    }
    while (c[i] > 0 && c[i] % 5 == 0) {
      b[i]++;
      c[i] /= 5;
    }
    sum1 += b[i];
    for (int j = sum1; j >= b[i]; j--) {
      for (int k = m; k >= 1; k--)
        if (dp[k - 1][j - b[i]] != 0)
          dp[k][j] = max(dp[k][j], dp[k - 1][j - b[i]] + a[i]);
    }
  }
  int ans = 0;
  for (int i = 0; i <= sum1; i++) {
    ans = max(ans, min(i, dp[m][i] - 1));
  }
  cout << ans << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  solve();
  return 0;
}
