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;
static char obuf[1 << 23], *p3 = obuf;
template <typename T>
inline void read(T& x) {
  T s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    s = (s << 1) + (s << 3) + (ch ^ 48), ch = getchar();
  }
  x = w * s;
}
template <typename T>
inline void write(T x) {
  if (x < 0) x = ~x + 1, *p3++ = '-';
  if (x > 9) write(x / 10);
  *p3++ = x % 10 + '0';
}
long long n, m, ans, a[205], _2[205], _5[205], dp[205][10005];
int main() {
  read(n), read(m);
  for (int i = 1; i <= n; ++i) {
    read(a[i]);
    while (a[i] % 2 == 0) {
      ++_2[i];
      a[i] >>= 1;
    }
    while (a[i] % 5 == 0) {
      ++_5[i];
      a[i] /= 5;
    }
  }
  memset(dp, -1, sizeof dp);
  dp[0][0] = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = m; j >= 0; j--) {
      for (int k = 10000; k >= 0; k--) {
        if (j >= 1 && k >= _5[i] && dp[j - 1][k - _5[i]] != -1)
          dp[j][k] = max(dp[j][k], dp[j - 1][k - _5[i]] + _2[i]);
      }
    }
  }
  for (int i = 0; i <= 10000; i++) {
    ans = max(ans, min((long long)i, dp[m][i]));
  }
  write(ans);
  fwrite(obuf, p3 - obuf, 1, stdout);
  return 0;
}
