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 = 202;
const int C = 6000;
int dp[N][C];
int n, k;
int a[N][2];
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    long long x;
    scanf("%lld", &x);
    while (x % 5 == 0) {
      a[i][0]++;
      x /= 5;
    }
    while (x % 2 == 0) {
      a[i][1]++;
      x /= 2;
    }
  }
  for (int i = 0; i <= k; i++)
    for (int j = 0; j < C; j++) dp[i][j] = -1;
  dp[0][0] = 0;
  for (int i = 0; i < n; i++)
    for (int j = k - 1; j >= 0; j--)
      for (int x = 0; x < C; x++) {
        if (dp[j][x] == -1) continue;
        dp[j + 1][x + a[i][0]] =
            max(dp[j + 1][x + a[i][0]], dp[j][x] + a[i][1]);
      }
  int ans = 0;
  for (int i = 0; i < C; i++) ans = max(ans, min(dp[k][i], i));
  printf("%d\n", ans);
  return 0;
}
