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 main() {
  int n, k, ans = 0;
  cin >> n >> k;
  int two[n], five[n];
  long long a;
  for (int i = 0; i < n; ++i) {
    cin >> a;
    five[i] = two[i] = 0;
    while (a && a % 5 == 0) {
      five[i]++;
      a /= 5;
    }
    while (a && a % 2 == 0) {
      two[i]++;
      a /= 2;
    }
  }
  int dp[2][n + 1][5200];
  for (int i = 0; i < 2; ++i) {
    for (int j = 0; j <= n; ++j) {
      for (int k = 0; k < 5200; ++k) {
        dp[i][j][k] = -20001;
      }
    }
  }
  for (int num_take = 1; num_take <= k; ++num_take) {
    for (int num_available = 1; num_available <= n; ++num_available) {
      for (int pow5 = 0; pow5 < 5200; ++pow5) {
        if (num_take > num_available) {
          dp[num_take % 2][num_available][pow5] = -20001;
          continue;
        }
        dp[num_take % 2][num_available][pow5] =
            max(dp[(num_take + 1) % 2][num_available][pow5],
                dp[num_take % 2][num_available - 1][pow5]);
        if (pow5 - five[num_available - 1] == 0) {
          dp[num_take % 2][num_available][pow5] = max(
              dp[num_take % 2][num_available][pow5], two[num_available - 1]);
        }
        if (pow5 - five[num_available - 1] >= 0) {
          dp[num_take % 2][num_available][pow5] =
              max(dp[num_take % 2][num_available][pow5],
                  dp[(num_take + 1) % 2][num_available - 1]
                    [pow5 - five[num_available - 1]] +
                      two[num_available - 1]);
        }
        ans = max(ans, min(pow5, dp[num_take % 2][num_available][pow5]));
      }
    }
  }
  cout << ans;
}
