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;
bool debug = false;
int dp[2][200][200];
int n, k;
int a[200];
int pw5(int num) {
  int cnt = 0;
  while (num % 5 == 0) {
    cnt++;
    num /= 5;
  }
  return cnt;
}
int pw2(int num) {
  int cnt = 0;
  while (num % 2 == 0) {
    num /= 2;
    cnt++;
  }
  return cnt;
}
int f[205][5556];
int main(int argc, char* argv[]) {
  cin >> n >> k;
  memset(f, -1, sizeof(f));
  f[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    long long a;
    cin >> a;
    int x = 0, y = 0;
    while (a % 2 == 0) a /= 2, x++;
    while (a % 5 == 0) a /= 5, y++;
    for (int j = k - 1; j >= 0; j--) {
      for (int l = 0; l < 5555; l++) {
        if (f[j][l] + 1) f[j + 1][l + y] = max(f[j + 1][l + y], f[j][l] + x);
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < 5555; i++) ans = max(ans, min(i, f[k][i]));
  cout << ans;
  return 0;
}
