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>
double a[100010];
inline int sign(double x) {
  if (x > 1e-8) return 1;
  if (x < -1e-8) return -1;
  return 0;
}
int main() {
  int n;
  scanf("%d", &n);
  bool flag = 1;
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    a[i] = x;
    if (!sign(a[i])) flag = 0;
  }
  if (flag) {
    bool now_flag = 1;
    double q = a[2] / a[1];
    for (int i = 3; i <= n; i++)
      if (sign(a[i] / a[i - 1] - q)) {
        now_flag = 0;
        break;
      }
    if (now_flag) {
      puts("0");
      return 0;
    }
  } else {
    int cnt = 0;
    for (int i = 1; i <= n; i++)
      if (!sign(a[i])) cnt++;
    if (cnt >= 2) {
      if (cnt == n)
        puts("0");
      else {
        if (a[1] != 0) cnt++;
        if (cnt == n)
          puts("0");
        else if (cnt == n - 1)
          puts("1");
        else
          puts("2");
      }
      return 0;
    }
    if (n == 1) {
      puts("0");
      return 0;
    }
    if (n == 2) {
      if (!sign(a[1]))
        puts("1");
      else
        puts("0");
      return 0;
    }
    for (int i = 1; i <= n; i++)
      if (!sign(a[i])) {
        for (int j = i; j < n; j++) a[j] = a[j + 1];
        break;
      }
    double q = a[2] / a[1];
    for (int i = 3; i < n; i++)
      if (sign(a[i] / a[i - 1] - q)) {
        puts("2");
        return 0;
      }
    puts("1");
    return 0;
    return 0;
  }
  flag = 1;
  double q = a[3] / a[2];
  for (int i = 4; i <= n; i++)
    if (sign(a[i] / a[i - 1] - q)) {
      flag = 0;
      break;
    }
  if (flag) {
    puts("1");
    return 0;
  }
  flag = 1;
  q = a[3] / a[1];
  for (int i = 4; i <= n; i++)
    if (sign(a[i] / a[i - 1] - q)) {
      flag = 0;
      break;
    }
  if (flag) {
    puts("1");
    return 0;
  }
  q = a[2] / a[1];
  for (int i = 3; i <= n; i++)
    if (fabs(a[i] / a[i - 1] - q) > 1e-8) {
      bool flag = 1;
      if (i != n && sign(a[i + 1] / a[i - 1] - q)) flag = 0;
      for (int j = i + 2; j <= n; j++)
        if (sign(a[j] / a[j - 1] - q)) flag = 0;
      if (!flag)
        puts("2");
      else
        puts("1");
      break;
    }
  return 0;
}
