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;
int n, k;
long long a[205];
int dp[205][22100];
int c2(long long x) {
  int ans = 0;
  while (x % 2 == 0) {
    ans++;
    x /= 2;
  }
  return ans;
}
int c5(long long x) {
  int ans = 0;
  while (x % 5 == 0) {
    ans++;
    x /= 5;
  }
  return ans;
}
int main() {
  scanf("%d%d", &n, &k);
  int big = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    big += c2(a[i]);
  }
  int x2, x5;
  for (int i = 0; i <= big; i++)
    for (int j = 0; j <= k; j++) dp[j][i] = -1;
  for (int i = 1; i <= n; i++) {
    x2 = c2(a[i]), x5 = c5(a[i]);
    for (int j = min(i, k); j >= 2; j--) {
      for (int l = 0; l <= big; l++) {
        if (l < x2) continue;
        if (dp[j - 1][l - x2] == -1) continue;
        dp[j][l] = max(dp[j][l], dp[j - 1][l - x2] + x5);
      }
    }
    dp[1][x2] = max(dp[1][x2], x5);
  }
  int ans = 0, t;
  for (int i = 0; i <= big; i++) {
    t = min(dp[k][i], i);
    ans = max(ans, t);
  }
  printf("%d\n", ans);
  return 0;
}
