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 long long MAXN = 2e2 + 69, MAX5 = 30, INF = 2e18;
struct vl {
  long long s2, s5;
};
long long n, k;
vl a[MAXN];
long long dp[MAXN][MAXN * MAX5];
void pt(long long i, long long x) {
  while (x % 2 == 0) {
    x /= 2;
    a[i].s2++;
  }
  while (x % 5 == 0) {
    x /= 5;
    a[i].s5++;
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (long long i = 1, x; i <= n; i++) cin >> x, pt(i, x);
  for (long long i = 0; i <= n; i++)
    for (long long j = 0; j <= n * MAX5; j++) dp[i][j] = -INF;
  dp[0][0] = 0;
  for (long long i = 1; i <= n; i++) {
    for (long long j = k; j >= 1; j--) {
      for (long long s5 = 0; s5 <= n * MAX5; s5++) {
        if (s5 < a[i].s5) continue;
        dp[j][s5] = max(dp[j][s5], dp[j - 1][s5 - a[i].s5] + a[i].s2);
      }
    }
  }
  long long ans = 0;
  for (long long s5 = 0; s5 <= n * MAX5; s5++)
    ans = max(ans, min(s5, dp[k][s5]));
  cout << ans;
}
