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 long long mod = 1e9 + 7;
const long long inf = (1ll << 30);
const int MX = 209;
long long dp[2][MX][MX * 30], n, k, x, f[MX], t[MX];
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &x);
    while (x % 2 == 0) {
      x /= 2;
      t[i]++;
    }
    while (x % 5 == 0) {
      x /= 5;
      f[i]++;
    }
  }
  long long cur = 1;
  for (int i = 0; i < 2; i++)
    for (int j = 0; j <= k; j++)
      for (int o = 0; o < MX * 30; o++) dp[i][j][o] = -inf;
  dp[0][0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= k; j++) {
      for (int o = 0; o < MX * 30; o++) {
        dp[cur][j][o] = dp[cur ^ 1][j][o];
        if (o - f[i] >= 0 && j - 1 >= 0)
          dp[cur][j][o] =
              max(dp[cur][j][o], dp[cur ^ 1][j - 1][o - f[i]] + t[i]);
      }
    }
    cur ^= 1;
  }
  long long ans = 0;
  for (long long i = 0; i < MX * 30; i++) {
    ans = max(ans, min(i, dp[cur ^ 1][k][i]));
  }
  cout << ans << endl;
}
