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;
mt19937_64 rng((long long)new char);
const double eps = 1e-8;
const int maxn = 200010;
const long double INF = 1e15;
const int mod = 1e9 + 7;
int dp[210][6010];
int c5[210], c2[210];
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, k;
  cin >> n >> k;
  for (int t = (1); t < (n + 1); ++t) {
    long long x;
    cin >> x;
    long long tx = x;
    c2[t] = 0;
    while (tx % 2 == 0) tx /= 2, ++c2[t];
    tx = x;
    c5[t] = 0;
    while (tx % 5 == 0) tx /= 5, ++c5[t];
  }
  memset(dp, 0xcf, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = k; j; --j) {
      for (int l = c5[i]; l < 6010; ++l) {
        dp[j][l] = max(dp[j][l], dp[j - 1][l - c5[i]] + c2[i]);
      }
    }
  }
  int ans = 0;
  for (int t = (0); t < (6010); ++t) ans = max(ans, min(t, dp[k][t]));
  cout << ans;
}
