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 a[210], b[210], dp[210][8100];
int ans, n, k;
int main() {
  scanf("%d%d", &n, &k);
  memset(dp, -1, sizeof(dp));
  for (int i = 1; i <= n; i++) {
    long long x;
    scanf("%lld", &x);
    while (x % 2 == 0) {
      a[i]++;
      x /= 2;
    }
    while (x % 5 == 0) {
      b[i]++;
      x /= 5;
    }
  }
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = k; j >= 1; j--) {
      for (int kk = 8000; kk >= b[i]; kk--) {
        if (dp[j - 1][kk - b[i]] != -1) {
          dp[j][kk] = max(dp[j][kk], dp[j - 1][kk - b[i]] + a[i]);
        }
      }
    }
  }
  for (int i = 0; i <= 8000; i++) {
    ans = max(ans, min(i, dp[k][i]));
  }
  printf("%d\n", ans);
  return 0;
}
