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;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int dp[210][5100];
int n, k;
int num1[210], num2[210];
int main() {
  n = read();
  k = read();
  for (int i = 1; i <= n; i++) {
    long long tmp = read();
    while (tmp % 2 == 0) {
      tmp /= 2;
      num1[i]++;
    }
    while (tmp % 5 == 0) {
      tmp /= 5;
      num2[i]++;
    }
  }
  memset(dp, 128, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = k - 1; j >= 0; j--) {
      for (int sum = 0; sum <= 4975; sum++) {
        dp[j + 1][sum + num2[i]] =
            max(dp[j + 1][sum + num2[i]], dp[j][sum] + num1[i]);
      }
    }
  }
  int maxx = 0;
  for (int i = 1; i <= 5000; i++) {
    int tmp = min(i, dp[k][i]);
    maxx = max(maxx, tmp);
  }
  cout << maxx << endl;
  return 0;
}
