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;
long long dp[500][6010];
long long a[234];
signed main() {
  for (long long i = 0; i <= 254; i++) {
    for (long long j = 0; j <= 6000; j++) {
      dp[i][j] = -1;
    }
  }
  dp[0][0] = 0;
  long long n, k;
  scanf("%I64d %I64d", &n, &k);
  for (long long i = 0; i < n; i++) {
    scanf("%I64d", &a[i]);
  }
  for (long long s = 0; s < n; s++) {
    long long g = a[s];
    long long t = 0, f = 0;
    while (g % 2 == 0) g /= 2, t++;
    g = a[s];
    while (g % 5 == 0) g /= 5, f++;
    for (long long i = k - 1; i >= 0; i--) {
      for (long long j = 0; j <= 6000; j++) {
        if (dp[i][j] != -1)
          dp[i + 1][j + f] = max(dp[i + 1][j + f], dp[i][j] + t);
      }
    }
  }
  long long Max = 0;
  for (long long i = 0; i <= 6000; i++) {
    Max = max(Max, min(i, dp[k][i]));
  }
  cout << Max << endl;
  return 0;
}
