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 MAXN = 205;
int f[2][MAXN][6005], sum2[MAXN], sum5[MAXN];
int main() {
  int N, K;
  scanf("%d%d", &N, &K);
  for (int i = 1; i <= N; i++) {
    long long x;
    cin >> x;
    while ((x & 1) == 0) {
      sum2[i]++;
      x >>= 1;
    }
    while (x % 5 == 0) {
      sum5[i]++;
      x /= 5;
    }
  }
  memset(f, -1, sizeof(f));
  f[0][0][0] = 0;
  for (int i = 1; i <= N; i++)
    for (int j = 0; j <= K; j++)
      for (int k = 0; k <= 6000; k++) {
        f[i % 2][j][k] = f[(i - 1) % 2][j][k];
        if (j >= 1 && k >= sum5[i] && f[(i + 1) % 2][j - 1][k - sum5[i]] != -1)
          f[i % 2][j][k] =
              max(f[i % 2][j][k], f[(i - 1) % 2][j - 1][k - sum5[i]] + sum2[i]);
      }
  int ans = 0;
  for (int k = 0; k < 6000; k++) ans = max(ans, min(k, f[N % 2][K][k]));
  printf("%d", ans);
  return 0;
}
