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;
double x[101010], b[101010], pre[101010], suc[101010];
int n, zero, m;
const double eps = 1e-6;
const int inf = 100000000;
int cmp(double x) {
  if (fabs(x) <= eps) return 0;
  if (x > eps)
    return 1;
  else
    return -1;
}
int check(int n) {
  if (n == 1) return 1;
  double B = b[2] / b[1];
  for (int i = 3; i <= n; i++)
    if (cmp(b[i] / b[i - 1] - B)) return 0;
  return 1;
}
int main() {
  scanf("%d", &n);
  zero = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lf", &x[i]);
    if (!cmp(x[i])) zero++;
  }
  if (zero == n) {
    cout << 0 << endl;
    return 0;
  }
  if (zero == n - 1 && cmp(x[1])) {
    cout << 0 << endl;
    return 0;
  }
  if (!zero) {
    for (int i = 1; i <= n; i++) b[i] = x[i];
    if (check(n)) {
      cout << 0 << endl;
      return 0;
    }
  }
  if (zero == n - 1) {
    cout << 1 << endl;
    return 0;
  }
  if (zero == n - 2 && cmp(x[1])) {
    cout << 1 << endl;
    return 0;
  }
  if (zero == 1) {
    m = 0;
    for (int i = 1; i <= n; i++)
      if (cmp(x[i])) b[++m] = x[i];
    if (check(m)) {
      cout << 1 << endl;
      return 0;
    }
  }
  if (zero == 0) {
    pre[1] = 0;
    for (int i = 2; i <= n; i++)
      if (!cmp(pre[i - 1] - inf))
        pre[i] = inf;
      else if (cmp(pre[i - 1]) && cmp(x[i] / x[i - 1] - pre[i - 1]))
        pre[i] = inf;
      else
        pre[i] = x[i] / x[i - 1];
    suc[n] = 0;
    for (int i = n - 1; i; i--)
      if (!cmp(suc[i + 1] - inf))
        suc[i] = inf;
      else if (cmp(suc[i + 1]) && cmp(x[i + 1] / x[i] - suc[i + 1]))
        suc[i] = inf;
      else
        suc[i] = x[i + 1] / x[i];
    int flag = 0;
    for (int i = 1; i <= n; i++)
      if (i == 1)
        flag |= (cmp(suc[i + 1] - inf));
      else if (i == n)
        flag |= (cmp(pre[i - 1] - inf));
      else {
        double a = pre[i - 1], b = suc[i + 1], c = x[i + 1] / x[i - 1];
        if (a == inf || b == inf) continue;
        if (cmp(a - b) && cmp(a) && cmp(b)) continue;
        if (cmp(a - c) && cmp(a) && cmp(c)) continue;
        if (cmp(b - c) && cmp(b) && cmp(c)) continue;
        flag = 1;
      }
    if (flag) {
      cout << 1 << endl;
      return 0;
    }
  }
  cout << 2 << endl;
  return 0;
}
