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 = 201;
int dp[2][N][5001];
int n, k;
int dois[N], cinco[N];
const int inf = int(1e9);
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    long long int x;
    cin >> x;
    while (x % 2 == 0) {
      x /= 2;
      dois[i]++;
    }
    while (x % 5 == 0) {
      x /= 5;
      cinco[i]++;
    }
  }
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < N; j++) {
      for (int k = 0; k < 5001; k++) {
        dp[i][j][k] = -inf;
      }
    }
  }
  dp[1][0][0] = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j <= k; j++) {
      for (int p5 = 0; p5 < 5001; p5++) {
        int on = i & 1;
        int ant = on ^ 1;
        dp[on][j][p5] = dp[ant][j][p5];
        if (j - 1 >= 0 && p5 - cinco[i] >= 0) {
          dp[on][j][p5] =
              max(dp[on][j][p5], dp[ant][j - 1][p5 - cinco[i]] + dois[i]);
        }
      }
    }
  }
  int ans = 0;
  for (int i = 1; i < 5001; i++) {
    ans = max(ans, min(i, dp[(n - 1) & 1][k][i]));
  }
  cout << ans;
  return 0;
}
