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;
int fiv[206], two[206];
int dp[201][32 * 201];
int main() {
  scanf("%d%d", &n, &k);
  int nfiv = 0, ntwo = 0;
  unsigned long long x;
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &x);
    while (!(x % 5)) {
      fiv[i]++;
      x /= 5;
      nfiv++;
    }
    while (!(x % 2)) {
      two[i]++;
      x /= 2;
      ntwo++;
    }
  }
  memset(dp, -1, sizeof(dp));
  for (int j = 0; j <= n; j++) dp[j][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = min(k, i); j >= 1; j--) {
      for (int w = nfiv; w >= fiv[i]; w--) {
        if (dp[j - 1][w - fiv[i]] != -1)
          dp[j][w] = max(dp[j][w], dp[j - 1][w - fiv[i]] + two[i]);
      }
    }
  }
  int ans = 0;
  for (int i = 1; i <= nfiv; i++) ans = max(ans, min(dp[k][i], i));
  printf("%d\n", ans);
  return 0;
}
