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;
char ch;
int bo;
inline bool blank(char ch) {
  return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
}
inline void RD(int &x) {
  x = bo = 0;
  for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') bo = 1;
  for (; ch >= '0' && ch <= '9';
       x = (x << 1) + (x << 3) + ch - '0', ch = getchar())
    ;
  if (bo) x = -x;
}
inline void RD(long long &x) {
  x = bo = 0;
  for (ch = getchar(); ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') bo = 1;
  for (; ch >= '0' && ch <= '9';
       x = (x << 1) + (x << 3) + ch - '0', ch = getchar())
    ;
  if (bo) x = -x;
}
inline long long RD() {
  long long x;
  RD(x);
  return x;
}
inline void RD(char *s) {
  for (ch = getchar(); blank(ch); ch = getchar())
    ;
  for (; !blank(ch); ch = getchar()) *s++ = ch;
  *s = 0;
}
inline void RD(char &c) {
  for (ch = getchar(); blank(c); ch = getchar())
    ;
}
template <class T>
inline void OT(T x) {
  static char buf[20];
  char *p1 = buf;
  if (!x) *p1++ = '0';
  if (x < 0) putchar('-'), x = -x;
  while (x) *p1++ = x % 10 + '0', x /= 10;
  while (p1-- != buf) putchar(*p1);
}
inline void pe() { puts(""); }
inline void pk() { putchar(' '); }
const double eps = 1e-8;
const int inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
const int maxn = 200 + 3;
int dp[maxn][27 * maxn];
long long a[maxn];
int main() {
  int n, k;
  RD(n), RD(k);
  for (int i = 0; i < n; i++) RD(a[i]);
  memset(dp, -1, sizeof(dp));
  int lim = 0;
  for (int i = 0; i < n; i++) {
    long long num = a[i];
    int n2 = 0, n5 = 0;
    while (num % 2 == 0) num /= 2, n2++;
    while (num % 5 == 0) num /= 5, n5++;
    for (int j = k - 1; j >= 1; j--) {
      for (int l = lim; l >= 0; l--)
        if (dp[j][l] != -1)
          dp[j + 1][l + n5] = max(dp[j + 1][l + n5], n2 + dp[j][l]);
    }
    lim += n5;
    dp[1][n5] = max(dp[1][n5], n2);
  }
  int ans = 0;
  for (int i = 0; i <= lim; i++) {
    ans = max(ans, min(i, dp[k][i]));
  }
  OT(ans), pe();
}
