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][202 * 28];
int cal(long long n, long long m) {
  int ret = 0;
  while (n % m == 0) {
    n /= m;
    ret++;
  }
  return ret;
}
int main() {
  int n, k, i, j, l, five, two, m;
  long long x;
  scanf("%d", &n);
  scanf("%d", &k);
  m = 28 * n;
  for (i = 0; i <= n; ++i) {
    for (j = 0; j <= m; ++j) {
      dp[0][i][j] = -1000000;
      dp[1][i][j] = -1000000;
    }
  }
  for (j = 0; j <= k; j++) {
    dp[0][j][0] = 0;
    dp[1][j][0] = 0;
  }
  for (i = 1; i <= n; ++i) {
    scanf("%lld", &x);
    five = cal(x, 5);
    two = cal(x, 2);
    for (j = 1; j <= k; ++j) {
      for (l = 0; l <= m; ++l) {
        if (l - five >= 0) {
          dp[i & 1][j][l] = max(dp[(i - 1) & 1][j - 1][l - five] + two,
                                dp[(i - 1) & 1][j][l]);
        } else {
          dp[i & 1][j][l] = dp[(i - 1) & 1][j][l];
        }
      }
    }
  }
  int ans = 0;
  for (i = 0; i <= m; ++i) {
    ans = max(ans, min(dp[n & 1][k][i], i));
  }
  printf("%d\n", ans);
}
