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 i, i1, j, k, k1, t, n, m, res, flag[10], a[210], b, dp[2][210][5210],
    p[210][2];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k1;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (i = 1; i <= n; i++) {
    p[i][0] = 0;
    p[i][1] = 0;
    while (a[i] % 2 == 0) {
      a[i] /= 2;
      p[i][0]++;
    }
    while (a[i] % 5 == 0) {
      a[i] /= 5;
      p[i][1]++;
    }
  }
  res = 0;
  for (i = 0; i < 2; i++) {
    for (j = 0; j <= n; j++) {
      for (k = 0; k < 5210; k++) {
        dp[i][j][k] = -(long long)1e18;
      }
    }
  }
  dp[0][0][0] = 0;
  for (i = 1; i <= n; i++) {
    for (j = 0; j <= n; j++) {
      for (k = 0; k < 5210; k++) {
        dp[1][j][k] = max(dp[1][j][k], dp[0][j][k]);
        if (k >= p[i][1] && j >= 1) {
          dp[1][j][k] = max(dp[1][j][k], dp[0][j - 1][k - p[i][1]] + p[i][0]);
        }
        if (j == k1) res = max(res, min(dp[1][j][k], k));
      }
    }
    for (j = 0; j <= n; j++) {
      for (k = 0; k < 5210; k++) {
        dp[0][j][k] = dp[1][j][k];
      }
    }
  }
  cout << res;
  return 0;
}
