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;
const int maxn = 3e5 + 5;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + 7;
long long quick_pow(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y & 1) {
      ans = ans * x % mod;
    }
    x = x * x % mod;
    y >>= 1;
  }
  return ans;
}
struct node {
  int cnt2;
  int cnt5;
} a[maxn];
int dp[205][205 * 64];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    long long x;
    scanf("%lld", &x);
    while (x % 2 == 0) {
      a[i].cnt2++;
      x /= 2;
    }
    while (x % 5 == 0) {
      a[i].cnt5++;
      x /= 5;
    }
  }
  long long sum = 0;
  memset(dp, -INF, sizeof(dp));
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    sum += a[i].cnt5;
    for (int j = min(k, i); j >= 1; j--) {
      for (int k = sum; k >= a[i].cnt5; k--) {
        dp[j][k] = max(dp[j][k], dp[j - 1][k - a[i].cnt5] + a[i].cnt2);
      }
    }
  }
  long long ans = 0;
  for (int i = 1; i <= sum; i++) {
    ans = max(ans, 1LL * min(i, dp[k][i]));
  }
  printf("%lld\n", ans);
  return 0;
}
