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 a[100005];
int n = 0;
double eps = 1e-6;
int ck(int i, double c) {
  double cur = a[i] * c;
  int cnt = 0;
  ++i;
  while (i < n) {
    if (fabs(a[i] - cur) > eps) {
      ++cnt;
      ++i;
    } else {
      cur = a[i] * c;
      ++i;
    }
    if (cnt >= 2) {
      break;
    }
  }
  return cnt;
}
int main() {
  memset(a, 0, sizeof(a));
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  if (n == 1) {
    printf("0");
  } else if (n == 2) {
    if (a[0] == 0 && a[1] != 0) {
      printf("1");
    } else {
      printf("0");
    }
  } else {
    double c = 0;
    int cnt = 0;
    if (a[0] == 0) {
      c = 0;
      if (a[1]) {
        ++cnt;
      }
    } else {
      c = (double)a[1] / a[0];
    }
    cnt += ck(1, c);
    if (cnt <= 1) {
      printf("%d", cnt);
      return 0;
    }
    if (a[1] == 0) {
      c = 0;
    } else {
      c = (double)a[2] / a[1];
    }
    cnt = min(cnt, ck(2, c) + 1);
    if (cnt <= 1) {
      printf("%d", cnt);
      return 0;
    }
    if (a[0] == 0) {
      c = 0;
    } else {
      c = (double)a[2] / a[0];
    }
    cnt = min(cnt, ck(2, c) + 1);
    printf("%d", cnt);
  }
  return 0;
}
