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, ans;
int dp[220][6500];
struct Node {
  long long num;
  int two, five;
} node[220];
void deal(int x) {
  long long now = node[x].num;
  while (now % 2 == 0) {
    now /= 2;
    node[x].two++;
  }
  while (now % 5 == 0) {
    now /= 5;
    node[x].five++;
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &node[i].num);
    deal(i);
  }
  memset(dp, -0x3f, sizeof dp);
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j >= 1; j--) {
      for (int p = 6200; p >= node[i].five; p--) {
        dp[j][p] = max(dp[j][p], dp[j - 1][p - node[i].five] + node[i].two);
      }
    }
  }
  for (int i = 1; i <= k; i++) {
    for (int j = 6200; j >= 1; j--) {
      ans = max(ans, min(j, dp[i][j]));
    }
  }
  printf("%d\n", ans);
  return 0;
}
