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 inf = 2147483647;
int two[205], five[205];
int dp[2][205][5200];
int possible[2][205][5200];
long long int a[205];
int main() {
  int n, K;
  scanf("%d%d", &n, &K);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", a + i);
    while (a[i] % 2 == 0) a[i] /= 2, two[i]++;
    while (a[i] % 5 == 0) a[i] /= 5, five[i]++;
  }
  dp[0][0][0] = 0;
  possible[0][0][0] = 1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 5200; j++) dp[i & 1][0][j] = dp[i & 1][i + 1][j] = 0;
    possible[i & 1][0][0] = 1;
    for (int j = 1; j <= i; j++) {
      for (int k = 0; k < 5200; k++) {
        dp[i & 1][j][k] = dp[1 - i & 1][j][k];
        possible[i & 1][j][k] = possible[1 - i & 1][j][k];
        if (possible[1 - i & 1][j - 1][k - five[i]]) possible[i & 1][j][k] = 1;
        if (k >= five[i] && possible[1 - i & 1][j - 1][k - five[i]]) {
          dp[i & 1][j][k] =
              max(dp[i & 1][j][k], dp[1 - i & 1][j - 1][k - five[i]] + two[i]);
        }
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < 5200; i++) {
    if (possible[n % 2][K][i]) ans = max(ans, min(i, dp[n % 2][K][i]));
  }
  printf("%d\n", ans);
}
