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;
vector<pair<int, int> > item;
int memo[2][202][5002];
pair<int, int> calc_power(unsigned long long value) {
  int two = 0, five = 0;
  while (value % 2 == 0) {
    two++;
    value /= 2;
  }
  while (value % 5 == 0) {
    five++;
    value /= 5;
  }
  return make_pair(two, five);
}
int mod(int val) { return val % 2; }
int dp() {
  memset(memo, -1, sizeof(memo));
  for (int i = (int)0; i <= (int)n; i++) {
    for (int j = (int)0; j <= (int)k; j++) {
      memo[mod(i)][j][0] = 0;
    }
  }
  for (int i = (int)1; i <= (int)n; i++) {
    for (int j = (int)1; j <= (int)k; j++) {
      for (int nfive = (int)0; nfive <= (int)5000; nfive++) {
        if (nfive >= item[i - 1].second) {
          int temp = memo[mod(i - 1)][j - 1][nfive - item[i - 1].second];
          if (temp > -1) {
            memo[mod(i)][j][nfive] =
                max(temp + item[i - 1].first, memo[mod(i)][j][nfive]);
          }
        }
        if (memo[mod(i - 1)][j][nfive] > -1) {
          memo[mod(i)][j][nfive] =
              max(memo[mod(i - 1)][j][nfive], memo[mod(i)][j][nfive]);
        }
      }
    }
  }
  int ans = 0;
  for (int i = (int)0; i <= (int)5000; i++) {
    ans = max(ans, min(i, memo[mod(n)][k][i]));
  }
  return ans;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < (int)n; i++) {
    unsigned long long temp;
    cin >> temp;
    item.push_back(calc_power(temp));
  }
  cout << dp() << endl;
}
