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;
#pragma comment(linker, "/STACK:20000000")
string problem_name = "a";
void init() { freopen((problem_name + ".in").c_str(), "rt", stdin); }
int mas[200100];
int dp[200100];
double dif;
int n;
int go(int pos) {
  if (dp[pos] != -1) return dp[pos];
  int res = 100;
  if (pos == n - 1) return 0;
  if (pos == n - 2) res = 1;
  if (abs(mas[pos] * dif - mas[pos + 1]) < 1e-9) res = min(res, go(pos + 1));
  if (pos < n - 1)
    if (abs(mas[pos] * dif - mas[pos + 2]) < 1e-9)
      res = min(res, 1 + go(pos + 2));
  return dp[pos] = res;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &mas[i]);
  double d = 0;
  int col = 0;
  for (int i = 1; i < n; i++) {
    if (mas[i] != mas[i - 1] * 0) col++;
  }
  if (col == 0) {
    printf("0");
    return 0;
  }
  col = 0;
  if (mas[0] != 0) {
    d = mas[1] * 1. / mas[0];
    for (int i = 0; i < n - 1; i++) {
      if (abs(mas[i] * d - mas[i + 1]) > 1e-9) col++;
    }
    if (col == 0) {
      printf("0");
      return 0;
    }
  }
  for (int i = 0; i <= n; i++) {
    if (i > 10) break;
    memset(dp, -1, sizeof(dp));
    if (mas[i] == 0 || i == n || i == 10)
      dif = 0;
    else
      dif = 1. * mas[i + 1] / mas[i];
    int res = go(0);
    if (res == 1) {
      printf("1\n");
      return 0;
    }
    memset(dp, -1, sizeof(dp));
    res = go(1);
    if (res == 0) {
      printf("1\n");
      return 0;
    }
  }
  printf("2");
  return 0;
}
