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 INF = 0x3f3f3f;
int n, k;
const int N = 206 * 64;
long long a[1000000];
long long dp[1001][N + 1];
int main() {
  scanf("%d%d", &n, &k);
  memset(dp, -INF, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);
  for (int i = 1; i <= n; i++) {
    long long n5 = 0, n2 = 0, q = a[i];
    while (q % 2 == 0) q /= 2, n2++;
    q = a[i];
    while (q % 5 == 0) q /= 5, n5++;
    for (int l = k; l >= 1; l--)
      for (int m = N; m >= n2; m--)
        dp[l][m] = max(dp[l][m], dp[l - 1][m - n2] + n5);
  }
  long long ans = 0;
  for (long long i = 1; i < N; i++) ans = max(ans, min(dp[k][i], i));
  cout << ans << endl;
  return 0;
}
