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;
pair<int, int> a[205];
long long dp[205][13105];
long long inf = 0x3f3f3f3f3f3f;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    long long v;
    scanf("%I64d", &v);
    int cnt = 0;
    while (v % 2 == 0) {
      v /= 2;
      cnt++;
    }
    a[i].first = cnt;
    cnt = 0;
    while (v % 5 == 0) {
      v /= 5;
      cnt++;
    }
    a[i].second = cnt;
  }
  for (int i = 0; i <= k; i++) {
    for (int j = 1; j <= 13000; j++) dp[i][j] = -inf;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = k; j >= 1; j--) {
      for (int l = 0; l + a[i].second <= 13000; l++) {
        dp[j][l + a[i].second] =
            max(dp[j][l + a[i].second], dp[j - 1][l] + a[i].first);
      }
    }
  }
  long long ans = 0;
  for (long long i = 0; i <= 13000; i++) {
    ans = max(ans, min(i, dp[k][i]));
  }
  printf("%I64d\n", ans);
}
