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;
const double EPS = 1e-8;
double s[100005];
double t[100005];
int pos;
int dmp(double k) {
  if (fabs(k) < EPS) return 0;
  return k < 0 ? -1 : 1;
}
int check(double num[], int n) {
  int i;
  if (n == 1) return 1;
  if (n == 2 && num[0] != 0) return 1;
  if (num[0] == 0) {
    for (int i = 0; i < n; i++)
      if (num[i] != 0) {
        pos = i;
        return 0;
      }
    return 1;
  }
  double b = num[1] / num[0];
  double temp = num[1];
  for (i = 2; i < n; i++) {
    temp *= b;
    if (dmp(temp - num[i]) != 0) {
      pos = i;
      return 0;
    }
  }
  return 1;
}
bool gao(int n) {
  int i, len;
  if (n == 2) return 1;
  for (i = 0; i < pos; i++) t[len++] = s[i];
  for (i = pos + 1; i < n; i++) t[len++] = s[i];
  if (check(t, n - 1)) return 1;
  for (i = 1; i < n; i++) t[i - 1] = s[i];
  if (check(t, n - 1)) return 1;
  t[0] = s[0];
  for (i = 2; i < n; i++) t[i - 1] = s[i];
  if (check(t, n - 1)) return 1;
  return 0;
}
int main() {
  int n;
  while (scanf("%d", &n) == 1) {
    for (int i = 0; i < n; i++) scanf("%lf", &s[i]);
    if (check(s, n)) {
      puts("0");
      continue;
    }
    if (gao(n))
      puts("1");
    else
      puts("2");
  }
  return 0;
}
