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;
struct node {
  long long t, f;
  node() { t = f = 0; }
  node(long long T, long long F) { t = T, f = F; }
} p[205];
long long n, k, a[205], dp[2][205][5005];
int main() {
  cin >> n >> k;
  for (long long i = 1; i <= n; ++i) {
    cin >> a[i];
  }
  for (long long i = 1; i <= n; ++i) {
    long long t = 0, f = 0;
    while (a[i] % 2 == 0) ++t, a[i] /= 2;
    while (a[i] % 5 == 0) ++f, a[i] /= 5;
    p[i] = node(t, f);
  }
  memset(dp, -1, sizeof dp);
  dp[0][0][0] = 0;
  long long up = 0;
  for (long long i = 1; i <= n; ++i) {
    long long now = i & 1;
    up += p[i].f;
    for (long long j = 0; j <= min(i, k); ++j) {
      for (long long l = 0; l <= up; ++l) {
        dp[now][j][l] = dp[now ^ 1][j][l];
        if (j && l >= p[i].f && ~dp[now ^ 1][j - 1][l - p[i].f])
          dp[now][j][l] =
              max(dp[now][j][l], dp[now ^ 1][j - 1][l - p[i].f] + p[i].t);
      }
    }
  }
  long long ans = 0;
  for (long long i = 0; i <= up; ++i) ans = max(ans, min(i, dp[n & 1][k][i]));
  cout << ans;
  return 0;
}
