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;
ifstream in("cf.in");
ofstream out("cf.out");
const int DIM = 1e2 + 5;
int dp[205][5205];
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie();
  cout.tie();
  int n, k;
  cin >> n >> k;
  memset(dp, -0x3f, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 1; i <= n; ++i) {
    long long x;
    int c2 = 0, c5 = 0;
    cin >> x;
    while (x % 2 == 0) {
      x /= 2;
      ++c2;
    }
    while (x % 5 == 0) {
      x /= 5;
      ++c5;
    }
    for (int i = k; i >= 1; --i)
      for (int j = c5; j < 5200; ++j)
        dp[i][j] = max(dp[i][j], dp[i - 1][j - c5] + c2);
  }
  int ans = 0;
  for (int j = 1; j <= 5200; ++j) ans = max(ans, min(j, dp[k][j]));
  cout << ans;
  return 0;
}
