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 n, i, a[100005], f[100005];
int g[100005], tag;
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  if (n == 1) {
    printf("0\n");
    return 0;
  }
  f[1] = 1;
  if (a[1] == 0 && a[2] != 0)
    f[2] = 0;
  else
    f[2] = 1;
  for (i = 3; i <= n; i++) {
    if (f[i - 1] == 0) break;
    if (a[i - 1] == 0) {
      if (a[i] == 0)
        f[i] = 1;
      else
        f[i] = 0;
    } else {
      if (a[i - 1] * a[i - 1] == a[i - 2] * a[i])
        f[i] = 1;
      else
        f[i] = 0;
    }
  }
  if (n == 2) {
    printf("%d\n", 1 - f[2]);
    return 0;
  }
  if (f[n] == 1) {
    printf("0\n");
    return 0;
  }
  if (f[n - 1] == 1) {
    printf("1\n");
    return 0;
  }
  g[n] = 1;
  if (a[n - 1] == 0 && a[n] != 0)
    g[n - 1] = 0;
  else
    g[n - 1] = 1;
  for (i = n - 2; i >= 1; i--) {
    if (g[i + 1] == 0) continue;
    if (a[i + 1] == 0) {
      if (a[i] == 0)
        g[i] = 1;
      else
        g[i] = 0;
    } else {
      if (a[i + 1] * a[i + 1] != a[i] * a[i + 2])
        g[i] = 0;
      else
        g[i] = 1;
    }
  }
  if (g[2] == 1) {
    printf("1\n");
    return 0;
  }
  if (n == 3) {
    printf("1\n");
    return 0;
  }
  for (i = 2; i < n; i++) {
    if (f[i - 1] == 0 || g[i + 1] == 0) continue;
    tag = 1;
    if (a[i - 1] == 0) {
      if (a[i + 1] == 0) {
        printf("1\n");
        return 0;
      } else
        tag = 0;
    }
    if (tag == 0) continue;
    if (i > 2) {
      if (a[i - 1] * a[i - 1] != a[i - 2] * a[i + 1]) tag = 0;
    }
    if (i < n - 1) {
      if (a[i + 1] * a[i + 1] != a[i - 1] * a[i + 2]) tag = 0;
    }
    if (tag == 1) {
      printf("1\n");
      return 0;
    }
  }
  printf("2\n");
  return 0;
}
