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 N = 205;
int n, _n, q, f[2][N][N * 30], s[N];
bool can[2][N][N * 30];
struct node {
  int cnt2, cnt5;
} a[N];
signed main() {
  scanf("%d %d", &n, &q);
  for (register int i = (1); i <= (n); ++i) {
    long long x;
    scanf("%lld", &x);
    while (x % 2 == 0) a[i].cnt2++, x /= 2;
    while (x % 5 == 0) a[i].cnt5++, x /= 5;
    s[i] = s[i - 1] + a[i].cnt5;
  }
  memset(f, -1, sizeof(f));
  f[0][0][0] = 0;
  for (register int i = (1); i <= (n); ++i) {
    memset(f[i & 1], -1, sizeof(f[i & 1]));
    f[i & 1][0][0] = 0;
    for (register int j = (1); j <= (min(q, i)); ++j)
      for (register int k = (0); k <= (s[i]); ++k) {
        f[i & 1][j][k] = f[(i - 1) & 1][j][k];
        if (j && k >= a[i].cnt5 && f[(i - 1) & 1][j - 1][k - a[i].cnt5] != -1)
          f[i & 1][j][k] = max(
              f[i & 1][j][k], f[(i - 1) & 1][j - 1][k - a[i].cnt5] + a[i].cnt2);
      }
  }
  int ans = 0;
  for (register int k = (1); k <= (s[n]); ++k)
    ans = max(ans, min(f[n & 1][q][k], k));
  printf("%d\n", ans);
  return 0;
}
