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;
int n, k;
short dp[2][201][201 * 26];
short two[201], five[201];
int main() {
  memset(dp, -1, sizeof dp);
  cin >> n >> k;
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    int cnt1 = 0, cnt2 = 0;
    long long u = x;
    while (u % 2 == 0) {
      cnt1++;
      u /= 2;
    }
    five[i] = cnt1;
    while (u % 5 == 0) {
      cnt2++;
      u /= 5;
    }
    two[i] = cnt2;
  }
  dp[0][k][0] = 0;
  int i = 0;
  for (int ii = 0; ii < n; ii++) {
    for (int j = 0; j <= k; j++) {
      for (int t = 0; t <= 201 * 26 - two[i] - 1; t++) {
        if (dp[i][j][t] == -1) continue;
        if (j != 0)
          dp[1 - i][j - 1][t + two[ii]] = max(dp[1 - i][j - 1][t + two[ii]],
                                              (short)(dp[i][j][t] + five[ii]));
        dp[1 - i][j][t] = max(dp[i][j][t], dp[1 - i][j][t]);
      }
    }
    i = 1 - i;
  }
  short ans = 0;
  for (int j = 0; j < 201 * 26; j++) ans = max(ans, min((short)j, dp[i][0][j]));
  cout << ans << endl;
  return 0;
}
