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;
template <typename T>
void chkmax(T &x, T y) {
  x = x > y ? x : y;
}
template <typename T>
void chkmin(T &x, T y) {
  x = x > y ? y : x;
}
const int INF = 2139062143;
template <typename T>
void read(T &x) {
  x = 0;
  bool f = 1;
  char ch;
  do {
    ch = getchar();
    if (ch == '-') f = 0;
  } while (ch > '9' || ch < '0');
  do {
    x = x * 10 + ch - '0';
    ch = getchar();
  } while (ch >= '0' && ch <= '9');
  x = f ? x : -x;
}
template <typename T>
void write(T x) {
  if (x < 0) x = ~x + 1, putchar('-');
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
const int N = 200 + 5;
const int M = 2e4 + 5;
long long n, K, P, ans, cnt2[N], cnt5[N], f[N][M];
signed main() {
  read(n);
  read(K);
  for (long long i = 1, x; i <= n; i++) {
    read(x);
    while (x % 2 == 0) cnt2[i]++, x /= 2;
    while (x % 5 == 0) cnt5[i]++, x /= 5;
    P += cnt5[i];
  }
  memset(f, -1, sizeof(f));
  f[0][0] = 0;
  for (long long i = 1; i <= n; i++) {
    for (long long j = K; j >= 1; j--) {
      for (long long k = P; k >= cnt5[i]; k--) {
        if (f[j - 1][k - cnt5[i]] == -1) continue;
        f[j][k] = max(f[j][k], f[j - 1][k - cnt5[i]] + cnt2[i]);
      }
    }
  }
  for (long long i = 0; i <= P; i++)
    if (~f[K][i]) ans = max(ans, min(f[K][i], i));
  cout << ans << endl;
  return 0;
}
