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, a[100010], b[100010], INF = 20000;
void f(int p, int q) {
  for (int i = (0); i < (100010); ++i) b[i] = INF;
  b[0] = p;
  b[1] = q;
  if (b[0] == 0) {
    for (int i = (0); i < (100010); ++i) b[i] = 0;
    return;
  }
  for (int i = (2); i < (100010); ++i) {
    if (abs(b[i - 1]) <= 10000 && abs(b[i - 1] * b[1]) % abs(b[0]) == 0 &&
        abs(b[i - 1] * b[1]) <= 10000 * abs(b[0])) {
      b[i] = (b[i - 1] * b[1]) / b[0];
    } else
      b[i] = INF;
  }
  return;
}
int main() {
  cin >> n;
  for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
  if (n == 1) {
    cout << "0" << endl;
    return 0;
  }
  if (n == 2) {
    if (a[0] == 0) {
      if (a[1] == 0)
        cout << "0" << endl;
      else
        cout << "1" << endl;
    } else
      cout << "0" << endl;
    return 0;
  }
  f(a[0], a[1]);
  int k1 = 0;
  for (int i = (0); i < (n); ++i)
    if (a[i] == b[k1]) k1++;
  f(a[1], a[2]);
  int k2 = 0;
  for (int i = (1); i < (n); ++i)
    if (a[i] == b[k2]) k2++;
  f(a[0], a[2]);
  int k3 = 0;
  for (int i = (0); i < (n); ++i) {
    if (i == 1) i++;
    if (a[i] == b[k3]) k3++;
  }
  if (k1 == n) {
    cout << "0" << endl;
    return 0;
  }
  if (k1 == n - 1) {
    cout << "1" << endl;
    return 0;
  }
  if (k2 == n - 1) {
    cout << "1" << endl;
    return 0;
  }
  if (k3 == n - 1) {
    cout << "1" << endl;
    return 0;
  }
  cout << "2" << endl;
  return 0;
}
