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;
int dp[5005][201][2] = {0}, a[201][2] = {0};
bool t[5005] = {1};
void f(long long b, int i) {
  while (b % 2 == 0) b /= 2, a[i][0]++;
  while (b % 5 == 0) b /= 5, a[i][1]++;
}
int main() {
  long long n, k, x;
  cin >> n >> k;
  dp[0][0][0] = 1;
  for (long long i = 0; i < n; ++i) {
    cin >> x, f(x, i);
    for (int j = min(i + 1, k); j; --j)
      for (int o = 5000; o >= a[i][1]; o--)
        if (dp[o - a[i][1]][j - 1][0])
          dp[o][j][1] = max(dp[o - a[i][1]][j - 1][1] + a[i][0], dp[o][j][1]),
          dp[o][j][0] = 1;
  }
  int m = 0;
  for (int i = 0; i <= 5000; ++i) m = max(min(dp[i][k][1], i), m);
  printf("%d", m);
}
