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 dp[2][202][28 * 204];
vector<pair<long long int, long long int> > A(202);
int n, k;
int main() {
  for (int i = 0; i < 2; i++)
    for (int j = 0; j < 202; j++)
      for (int l = 0; l < 202 * 28; l++) dp[i][j][l] = -10000000;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    long long int temp;
    cin >> temp;
    int two = 0, five = 0;
    while (temp % 2 == 0) temp = temp / 2, two++;
    while (temp % 5 == 0) temp = temp / 5, five++;
    A[i].first = two;
    A[i].second = five;
  }
  dp[0][0][0] = 0;
  for (int i = 1; i <= n; i++) {
    int two = A[i].first;
    int five = A[i].second;
    for (int j = min(k, i); j >= 0; j--) {
      for (int l = 0; l < 202 * 28; l++) {
        if (l >= five && j > 0)
          dp[i & 1][j][l] =
              max(dp[i & 1][j][l], dp[~i & 1][j - 1][l - five] + two);
        dp[i & 1][j][l] = max(dp[i & 1][j][l], dp[~i & 1][j][l]);
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < 202 * 28; i++) {
    ans = max(ans, min(i, dp[n & 1][k][i]));
  }
  cout << ans;
  return 0;
}
