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 long long MXN = 205;
const long long MX5 = MXN * 27;
long long n, k;
long long dp[MXN][MX5];
int main() {
  memset(dp, ~0x3f, sizeof(dp));
  dp[0][0] = 0;
  scanf("%lld%lld", &n, &k);
  for (long long i = 1; i <= n; i++) {
    long long tmp, c2 = 0, c5 = 0;
    scanf("%lld", &tmp);
    while (!(tmp & 1)) {
      tmp >>= 1;
      c2++;
    }
    while (!(tmp % 5)) {
      tmp /= 5;
      c5++;
    }
    for (long long j = min(i, k); j; j--)
      for (long long l = MX5 - 1; l >= c5; l--)
        dp[j][l] = max(dp[j][l], dp[j - 1][l - c5] + c2);
  }
  long long ans = 0;
  for (long long i = 1; i < MX5; i++) ans = max(ans, min(i, dp[k][i]));
  printf("%lld", ans);
  return 0;
}
