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;
const int N = 205;
ll dp[2][N][5305], p2[N], p5[N], ans, num;
int n, k;
ll calc(ll x, ll y) {
  ll cnt = 0;
  while (x % y == 0) {
    x /= y;
    ++cnt;
  }
  return cnt;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    scanf("%lld", &num);
    p2[i] = calc(num, 2ll);
    p5[i] = calc(num, 5ll);
  }
  memset(dp, 0xcf, sizeof dp);
  dp[0][0][0] = 0;
  for (int i = 1; i <= n; ++i) {
    for (int j = 0; j <= k; ++j) {
      for (int _ = 0; _ <= 5300; ++_) {
        if (j > 0 && _ >= p5[i])
          dp[i % 2][j][_] = max(dp[(i - 1) % 2][j - 1][_ - p5[i]] + p2[i],
                                dp[(i - 1) % 2][j][_]);
        else
          dp[i % 2][j][_] = dp[(i - 1) % 2][j][_];
      }
    }
  }
  for (int i = 0; i <= 5300; ++i) {
    ans = max(ans, min(dp[n % 2][k][i], (ll)i));
  }
  cout << ans << endl;
}
