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 get2(long long x) {
  int ret = 0;
  while (x % 2 == 0) {
    ret++;
    x /= 2;
  }
  return ret;
}
int get5(long long x) {
  int ret = 0;
  while (x % 5 == 0) {
    ret++;
    x /= 5;
  }
  return ret;
}
const int maxn = 64 * 210;
int dp[210][maxn];
int main() {
  int n, m;
  long long t;
  scanf("%d%d", &n, &m);
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j < maxn; j++) {
      dp[i][j] = -1000000000;
    }
  }
  dp[0][0] = 0;
  for (int k = 1; k <= n; k++) {
    scanf("%lld", &t);
    int n2 = get2(t);
    int n5 = get5(t);
    for (int i = k; i > 0; i--) {
      for (int j = n2; j < maxn; j++) {
        dp[i][j] = max(dp[i][j], dp[i - 1][j - n2] + n5);
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < maxn; i++) {
    ans = max(ans, min(i, dp[m][i]));
  }
  printf("%d\n", ans);
  return 0;
}
