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 = 200 + 1;
int p[2][maxn];
int d[2][maxn][5001], sum[maxn];
void upd(int& a, int b) {
  if (b > a) a = b;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i < n + 1; i++) {
    long long x;
    scanf("%lld", &x);
    while (!(x & 1)) {
      p[0][i]++;
      x >>= 1;
    }
    while (x % 5 == 0) {
      p[1][i]++;
      x /= 5;
    }
  }
  for (int i = 1; i < n + 1; i++) sum[i] = sum[i - 1] + p[1][i];
  memset(d, -1, sizeof(d));
  d[1][0][0] = 0;
  d[1][1][p[1][1]] = p[0][1];
  for (int i = 1; i < n; i++) {
    memset(d[(i & 1) ^ 1], -1, sizeof(d[0]));
    for (int j = 0; j < k + 1; j++)
      for (int l = 0; l < sum[i] + 1; l++)
        if (d[i & 1][j][l] >= 0) {
          int& t = d[i & 1][j][l];
          upd(d[(i & 1) ^ 1][j][l], t);
          upd(d[(i & 1) ^ 1][j + 1][l + p[1][i + 1]], t + p[0][i + 1]);
        }
  }
  int ans = 0;
  for (int i = 1; i < 5001; i++)
    if (d[n & 1][k][i] >= 0) ans = max(ans, min(i, d[n & 1][k][i]));
  printf("%d\n", ans);
  return 0;
}
