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>
#pragma GCC target("popcnt")
using namespace std;
const int mxN = 201;
const int mxlog5 = 30;
const double neginf = -1e18;
pair<int, int> a[mxN];
int dp[mxN][mxN * mxlog5];
int dp_new[mxN][mxN * mxlog5];
int main() {
  int n, k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    long long x;
    cin >> x;
    while (x % 2 == 0) x /= 2, a[i].first++;
    while (x % 5 == 0) x /= 5, a[i].second++;
  }
  memset(dp, -1, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    memset(dp_new, -1, sizeof(dp_new));
    int two = a[i - 1].first;
    int five = a[i - 1].second;
    for (int j = 1; j <= k; j++) {
      for (int o = 0; o + five < mxN * mxlog5; o++) {
        if (dp[j - 1][o] != -1)
          dp_new[j][o + five] = max(dp_new[j][o + five], dp[j - 1][o] + two);
        if (dp[j][o] != -1) dp_new[j][o] = max(dp_new[j][o], dp[j][o]);
      }
    }
    for (int j = 1; j <= k; j++) {
      for (int o = 0; o < mxN * mxlog5; o++) {
        dp[j][o] = dp_new[j][o];
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < mxN * mxlog5; i++) {
    ans = max(ans, min(dp[k][i], i));
  }
  cout << ans;
}
