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;
int dp[5250][210];
long long tv[250], fv[250];
int main() {
  for (int i = 0; i < 5250; i++)
    for (int j = 0; j < 210; j++) {
      dp[i][j] = -999999;
    }
  dp[0][0] = 0;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    long long t;
    cin >> t;
    int c2 = 0, c5 = 0;
    while (!(t % 2)) {
      t /= 2;
      c2++;
    }
    while (!(t % 5)) {
      t /= 5;
      c5++;
    }
    tv[i] = c2;
    fv[i] = c5;
  }
  for (int i = 0; i < n; i++) {
    int it = min(k, i + 1);
    for (int j = it - 1; j >= 0; j--) {
      for (int l = 0; l < 26 * (i + 1); l++) {
        dp[l + fv[i]][j + 1] = max(dp[l + fv[i]][j + 1], dp[l][j] + (int)tv[i]);
      }
    }
  }
  int mp = 0;
  for (int i = 0; i < 5250; i++) {
    mp = max(mp, min(dp[i][k], i));
  }
  cout << mp << '\n';
}
