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 n, k;
long long x;
int dp[200 * 26][205], c[205], w[205];
int maxv = 0;
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &x);
    long long t = x;
    int n2 = 0, n5 = 0;
    while (t % 2 == 0) {
      n2++;
      t /= 2;
    }
    while (t % 5 == 0) {
      n5++;
      t /= 5;
    }
    w[i] = n5;
    c[i] = n2;
    maxv += n5;
  }
  memset(dp, -0x3f3f3f3f, sizeof dp);
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++)
    for (int v = maxv; v >= 0; v--)
      for (int t = k; t >= 1; t--)
        if (v - w[i] >= 0) dp[v][t] = max(dp[v][t], dp[v - w[i]][t - 1] + c[i]);
  int res = 0;
  for (int v = maxv; v >= 0; v--) res = max(res, min(v, dp[v][k]));
  printf("%d\n", res);
  return 0;
}
