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;
const long long inf = (long long)1e14;
const int mod = (int)1e9 + 7;
const int N = (int)100100;
const double eps = 1e-10;
const double pi = acos(-1.0);
long long a[N], dp[2][201][5300], pw5[201], pw2[201];
long long five(long long x) {
  long long cnt = 0;
  while (x % 5 == 0) {
    cnt++;
    x /= 5;
  }
  return cnt;
}
long long two(long long x) {
  long long cnt = 0;
  while (x % 2 == 0) {
    cnt++;
    x /= 2;
  }
  return cnt;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    pw2[i] = two(a[i]);
    pw5[i] = five(a[i]);
  }
  for (int i = 0; i <= 1; i++) {
    for (int j = 0; j <= 200; j++) {
      for (int t = 0; t <= 5200; t++) {
        dp[i][j][t] = -inf;
      }
    }
  }
  dp[1][1][pw5[1]] = pw2[1];
  dp[1][0][0] = 0;
  for (int i = 2; i <= n; i++) {
    for (int j = 0; j <= min((long long)i, k); j++) {
      for (int t = 0; t <= 5200; t++) {
        int cur = (i & 1);
        int prev = ((i - 1) & 1);
        if (t - pw5[i] >= 0 && j > 0) {
          dp[cur][j][t] =
              max(dp[cur][j][t], dp[prev][j - 1][t - pw5[i]] + pw2[i]);
        }
        dp[cur][j][t] = max(dp[cur][j][t], dp[prev][j][t]);
      }
    }
    for (int j = 0; j <= k; j++) {
      for (int t = 0; t <= 5200; t++) {
        int prev = ((i - 1) & 1);
        dp[prev][j][t] = -inf;
      }
    }
  }
  int cur = (n & 1);
  long long ans = 0;
  for (int i = 0; i <= 5200; i++) {
    long long deg = min((long long)i, dp[cur][k][i]);
    ans = max(ans, deg);
  }
  cout << ans;
  return 0;
}
