Polycarp loves geometric progressions — he collects them. However, as such progressions occur very rarely, he also loves the sequences of numbers where it is enough to delete a single element to get a geometric progression.

In this task we shall define geometric progressions as finite sequences of numbers a1, a2, ..., ak, where ai = c·bi - 1 for some real numbers c and b. For example, the sequences [2, -4, 8], [0, 0, 0, 0], [199] are geometric progressions and [0, 1, 2, 3] is not.

Recently Polycarp has found a sequence and he can't classify it. Help him to do it. Determine whether it is a geometric progression. If it is not, check if it can become a geometric progression if an element is deleted from it.

Input

The first line contains an integer n (1 ≤ n ≤ 105) — the number of elements in the given sequence. The second line contains the given sequence. The numbers are space-separated. All the elements of the given sequence are integers and their absolute value does not exceed 104.

Output

Print 0, if the given sequence is a geometric progression. Otherwise, check if it is possible to make the sequence a geometric progression by deleting a single element. If it is possible, print 1. If it is impossible, print 2.

Examples

Input

4
3 6 12 24


Output

0


Input

4
-8 -16 24 -32


Output

1


Input

4
0 1 2 3


Output

2
#include <bits/stdc++.h>
using namespace std;
int i, j, m, n, p, k, a[200005], sum, flag, b[200005];
inline int read() {
  int x, f;
  char c;
  while (c = getchar(), (c < '0' || c > '9') && c != '-')
    ;
  if (c == '-')
    f = -1, c = getchar();
  else
    f = 1;
  x = c - '0';
  while (c = getchar(), (c >= '0' && c <= '9'))
    x = (x << 1) + (x << 3) + c - '0';
  return x * f;
}
void li() {
  if (flag) return;
  puts("0");
  flag = 1;
}
void yi() {
  if (flag) return;
  puts("1");
  flag = 1;
}
void er() {
  if (flag) return;
  puts("2");
  flag = 1;
}
bool del(int x) {
  int i, j, k;
  for (i = 1; i <= n; ++i) b[i] = a[i];
  for (i = x + 1; i <= n; ++i) b[i - 1] = b[i];
  --n;
  for (i = 2; i < n; ++i)
    if (b[i] * b[i] != b[i - 1] * b[i + 1]) break;
  ++n;
  if (i == n - 1) return 1;
  return 0;
}
int main() {
  while (scanf("%d", &n) != EOF) {
    sum = flag = 0;
    for (i = 1; i <= n; ++i) {
      a[i] = read();
      if (!a[i]) ++sum;
    }
    if (n == 1 || sum == n) li();
    if (sum == n - 1 && a[1] != 0) li();
    if (sum == 0 && n == 2) li();
    if (sum == n - 1) yi();
    if (sum == n - 2 && a[1] != 0) yi();
    if (sum == 1) {
      for (i = 1; i <= n; ++i)
        if (!a[i]) break;
      if (del(i))
        yi();
      else
        er();
    }
    if (sum) er();
    for (i = 2; i < n; ++i)
      if (a[i] * a[i] != a[i - 1] * a[i + 1]) break;
    if (i == n) li();
    for (j = i - 1; j <= i + 1; ++j)
      if (del(j)) yi();
    er();
  }
}
