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;
void re() { vector<int> b(1e10); }
const int N = 209;
const int F = 26;
int dp[2][N][F * N];
void get25(long long n, int &two, int &five) {
  while (n % 2LL == 0) {
    n /= 2LL;
    two++;
  }
  while (n % 5LL == 0) {
    n /= 5LL;
    five++;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, K;
  cin >> n >> K;
  vector<long long> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  int p = 0;
  int c = 1;
  dp[p][0][0] = 1;
  int ans = 0;
  for (int i = 0; i < n; ++i, swap(p, c)) {
    for (int k = 0; k < N; ++k)
      for (int f = 0; f < F * N; ++f) dp[c][k][f] = 0;
    int two = 0;
    int five = 0;
    get25(a[i], two, five);
    for (int k = 0; k <= K; ++k) {
      for (int f = 0; f < F * N; ++f) {
        dp[c][k][f] = dp[p][k][f];
        if (f - five >= 0 && dp[p][k - 1][f - five] && k - 1 >= 0) {
          dp[c][k][f] = max(dp[c][k][f], dp[p][k - 1][f - five] + two);
        }
        if (dp[c][k][f]) {
          ans = max(ans, min(f, dp[c][k][f] - 1));
        }
      }
    }
  }
  cout << ans;
}
