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, i, j, l, mx, shi;
long long a[201];
int zhi2[201], zhi5[201];
int dp[201][10001];
int zhi(long long x, int y) {
  int cnt = 0;
  while (x % y == 0) {
    cnt++;
    x = x / y;
  }
  return cnt;
}
int main() {
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    zhi2[i] = zhi(a[i], 2);
    zhi5[i] = zhi(a[i], 5);
  }
  for (i = 0; i <= 200; i++)
    for (j = 0; j <= 10000; j++) dp[i][j] = -1;
  dp[0][0] = 0;
  for (i = 1; i <= n; i++) {
    for (j = k; j >= 1; j--)
      for (l = 10000; l >= zhi5[i]; l--) {
        if (dp[j - 1][l - zhi5[i]] != -1) {
          dp[j][l] = max(dp[j - 1][l - zhi5[i]] + zhi2[i], dp[j][l]);
        }
      }
  }
  mx = 0;
  for (j = 0; j <= 10000; j++) {
    shi = min(dp[k][j], j);
    mx = max(mx, shi);
  }
  cout << mx;
  return 0;
}
