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 MN = 201, MP = 5001;
pair<int, int> in[MN];
int dp[2][MN][MP];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    long long a;
    scanf("%lld", &a);
    while (a % 2 == 0) {
      in[i].first++;
      a /= 2;
    }
    while (a % 5 == 0) {
      in[i].second++;
      a /= 5;
    }
  }
  for (int i = 0; i < MP; ++i)
    for (int k = 0; k < MN; ++k) dp[0][k][i] = -1e7;
  dp[0][0][0] = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = 0; j <= k; ++j)
      for (int b = 0; b < MP; ++b) {
        dp[i % 2][j][b] = dp[(i - 1) % 2][j][b];
        if (j >= 1 && b >= in[i].second) {
          dp[i % 2][j][b] =
              max(dp[i % 2][j][b],
                  dp[(i - 1) % 2][j - 1][b - in[i].second] + in[i].first);
        }
      }
  int ans = 0;
  for (int i = 0; i < MP; ++i) {
    ans = max(ans, min(i, dp[n % 2][k][i]));
  }
  printf("%d\n", ans);
}
