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 dx[4] = {0, 0, 1, -1};
int dy[4] = {1, -1, 0, 0};
const double eps = 1e-8;
const int mod = 1000000007;
const double PI = acos(-1.0);
inline int read() {
  int num;
  scanf("%d", &num);
  return num;
}
const int maxn = 1007;
int dp[207][6007];
int two[207];
int five[207];
void solve(int i, long long num) {
  while (num != 0 && num % 5 == 0) {
    five[i]++;
    num /= 5;
  }
  while (num != 0 && num % 2 == 0) {
    two[i]++;
    num /= 2;
  }
}
int main() {
  int n = read();
  int m = read();
  for (int i = 1; i <= n; i++) {
    long long num;
    cin >> num;
    solve(i, num);
  }
  memset(dp, -1, sizeof dp);
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i; j >= 0; j--) {
      for (int k = 6000; k >= five[i]; k--) {
        if (k >= five[i] && ~dp[j - 1][k - five[i]] && j >= 1)
          dp[j][k] = max(dp[j][k], dp[j - 1][k - five[i]] + two[i]);
      }
    }
  }
  int ans = 0;
  for (int i = 0; i <= 6000; i++) {
    ans = max(min(dp[m][i], i), ans);
  }
  cout << ans << endl;
  return 0;
}
