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;
using ll = long long;
using lb = long double;
using db = double;
const ll mod = 1e9 + 7;
const int maxn = 1000005;
const int INF = 10000000;
int dp[205][20000];
int main() {
  int n, k;
  cin >> n >> k;
  ll a;
  int sum = 0;
  for (int i = 0; i < 205; i++)
    for (int j = 0; j < 205 * 60; j++) {
      dp[i][j] = -INF;
    }
  dp[0][0] = 0;
  for (int s = 1; s <= n; s++) {
    scanf("%lld", &a);
    ll temp = a;
    int x = 0, y = 0;
    while (temp % 2 == 0 && temp != 0) {
      temp /= 2;
      x++;
    }
    while (temp % 5 == 0 && temp != 0) {
      temp /= 5;
      y++;
    }
    sum += x;
    for (int i = min(s, k); i >= 1; i--) {
      for (int j = x; j <= sum; j++) {
        dp[i][j] = max(dp[i][j], dp[i - 1][j - x] + y);
      }
    }
  }
  ll ans = 0, as = 0;
  for (int i = 0; i <= sum; i++) {
    ans = min(i, dp[k][i]);
    as = max(as, ans);
  }
  cout << as << endl;
  return 0;
}
