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;
long long a[N], t[N], f[N];
long long dp[N][N * 25];
int main() {
  int n, k;
  long long w, b = 0;
  scanf("%d%d", &n, &k);
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    long long w = a[i];
    f[i] = t[i] = 0;
    while (w % 2 == 0) {
      t[i]++;
      w /= 2;
    }
    while (w % 5 == 0) {
      f[i]++;
      w /= 5;
    }
    b += f[i];
    for (int j = min(i, k); j >= 1; j--) {
      for (int k = b; k >= f[i]; k--) {
        if (dp[j - 1][k - f[i]] == -1) continue;
        dp[j][k] = max(dp[j][k], dp[j - 1][k - f[i]] + t[i]);
      }
    }
  }
  long long res = 0;
  for (int j = 1; j <= b; j++) {
    res = max(res, min((long long)j, dp[k][j]));
  }
  printf("%lld\n", res);
}
