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 = 0.000000001;
double x[200000];
int n;
void solve() {
  if (abs(x[1]) < eps) {
    int p = 0;
    for (int i = 1; i < n; i++)
      if (abs(x[i]) > eps) p++;
    if (p == 0) {
      printf("1\n");
      exit(0);
    }
    return;
  }
  double q = x[2] / x[1];
  for (int i = 3; i < n; i++) {
    if (abs(x[i]) < eps) {
      if (abs(q) > eps) return;
    } else if (abs(x[i] / x[i - 1] - q) > eps)
      return;
  }
  printf("1\n");
  exit(0);
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) scanf("%lf", &x[i]);
  if (abs(x[0]) < eps) {
    int p = 0;
    for (int i = 0; i < n; i++)
      if (abs(x[i]) > eps) p++;
    if (p < 2) {
      printf("%d\n", p);
      return 0;
    }
    solve();
    printf("2\n");
    return 0;
  }
  if (n < 3) {
    printf("0\n");
    return 0;
  }
  double q = x[1] / x[0];
  int p = 0;
  for (int i = 2; i < n; i++) {
    if (abs(x[i]) < eps) {
      if (abs(q) > eps) {
        if (p == 1) {
          p = 2;
          break;
        }
        p = 1;
        x[i] = x[i - 1];
      }
    } else if (abs(x[i] / x[i - 1] - q) > eps) {
      if (p == 1) {
        p = 2;
        break;
      }
      p = 1;
      if (abs(x[i] - x[i - 1]) > eps && i != n - 1) {
        if (abs(x[i + 1] / x[i - 1] - q) < eps)
          x[i] = x[i - 1];
        else if (abs(x[i + 1] / x[i] - q) > eps) {
          p = 2;
          break;
        }
      }
    }
  }
  if (p < 2) {
    printf("%d\n", p);
    return 0;
  }
  solve();
  swap(x[0], x[1]);
  solve();
  printf("2\n");
  return 0;
}
