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 const pi = 3.1415926535897932384626433832795;
int const inf = (int)1e9;
long long const inf64 = (long long)2e18;
int n, a[200001], p[200001];
vector<double> b;
bool eq(double x, double y) {
  if (fabs(x - y) > 1e-8)
    return false;
  else
    return true;
}
bool check(int n) {
  if (n <= 2) {
    if (n == 2 && p[0] == 0 && p[1] == 0 || n == 1 || p[0] != 0)
      return true;
    else
      return false;
  }
  double B;
  if (p[0] == 0)
    B = 0;
  else
    B = (double)p[1] / p[0];
  for (int i = 1; i < n; i++)
    if (!eq(p[i], p[i - 1] * B)) return false;
  return true;
}
int main() {
  cin >> n;
  for (int i = 0; i < (int)n; i++) scanf("%d", &a[i]);
  for (int i = 0; i < (int)n; i++) p[i] = a[i];
  if (check(n)) {
    cout << 0 << endl;
    return 0;
  }
  if (n < 100) {
    for (int i = 0; i < (int)n; i++) {
      int now = 0;
      for (int j = 0; j < (int)n; j++)
        if (j != i) p[now++] = a[j];
      if (check(n - 1)) {
        cout << 1 << endl;
        return 0;
      }
    }
    cout << 2 << endl;
    return 0;
  }
  for (int i = 0; i < (int)n - 1; i++) p[i] = a[i + 1];
  if (check(n - 1)) {
    cout << 1 << endl;
    return 0;
  }
  {
    int cnt = 0;
    for (int i = 0; i < (int)n; i++)
      if (i && a[i] == 0) cnt++;
    if (cnt > 1 && cnt < n - 2) {
      cout << 2 << endl;
      return 0;
    }
    if (cnt == n - 2) {
      cout << 1 << endl;
      return 0;
    }
  }
  for (int i = 0; i < (int)n; i++)
    if (a[i] == 0) {
      int now = 0;
      for (int j = 0; j < (int)n; j++)
        if (j != i) p[now++] = a[j];
      if (check(n - 1))
        cout << 1 << endl;
      else
        cout << 2 << endl;
      return 0;
    }
  for (int i = 0; i < (int)n - 1; i++) p[i] = a[i];
  if (check(n - 1)) {
    cout << 1 << endl;
    return 0;
  }
  for (int i = 0; i < (int)n - 1; i++) b.push_back((double)a[i] / a[i + 1]);
  sort((b).begin(), (b).end());
  int cnt = 0, best = 0;
  double B = 0;
  sort((b).begin(), (b).end());
  b.push_back(b[0] - 1);
  for (int i = 0; i < (int)b.size(); i++)
    if (!i || !eq(b[i], b[i - 1])) {
      if (i && best < cnt) {
        best = cnt;
        B = b[i - 1];
      }
      cnt = 1;
    } else
      cnt++;
  for (int i = 1; i < n - 1; i++) {
    int now = best;
    if (eq((double)a[i] / a[i + 1], B)) now--;
    if (eq((double)a[i - 1] / a[i], B)) now--;
    if (eq((double)a[i - 1] / a[i + 1], B)) now++;
    if (now == n - 2) {
      cout << 1 << endl;
      return 0;
    }
  }
  cout << 2 << endl;
  return 0;
}
