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>
int a[100010], b[100010];
bool check(int *a, int n, int id) {
  int cntt = 0;
  for (int i = 0; i < n; i++) {
    if (i == id) continue;
    b[cntt++] = a[i];
  }
  if (cntt == 1) return true;
  for (int i = 1; i < cntt; i++)
    if (b[i - 1] == 0 && b[i] != 0) return false;
  if (cntt == 2) return true;
  for (int i = 1; i < cntt - 1; i++) {
    int num = b[i] * b[i];
    if (num != b[i - 1] * b[i + 1]) return false;
  }
  return true;
}
bool check2(int *aa, int n) {
  for (int i = 0; i < n; i++) b[i] = aa[i];
  int cntt = 0;
  if (b[0] == 0 && b[1] != 0) return false;
  if (b[1] == 0 && b[2] != 0) return false;
  if (b[1] * b[1] != b[0] * b[2]) return false;
  bool zero = true;
  int zeros = 0;
  for (int i = 0; i < 3; i++) {
    if (b[i] != 0)
      zero = false;
    else
      zeros++;
  }
  if (zeros != 0) {
    int cntt = 0;
    for (int i = 3; i < n; i++)
      if (b[i] != 0) cntt++;
    if (cntt == 1) return true;
    return false;
  }
  for (int i = 2; i < n; i++) {
    int next = b[i - 1] * b[i - 1];
    if (next % b[i - 2] != 0) return false;
    next = next / b[i - 2];
    if (next != b[i]) {
      cntt++;
      if (i != n - 1 && next != b[i + 1]) return false;
      i++;
      b[i - 1] = b[i - 2];
    }
  }
  if (cntt == 1) return true;
  return false;
}
int main() {
  int n;
  while (scanf("%d", &n) == 1) {
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    int zeros = 0;
    for (int i = 0; i < n; i++)
      if (a[i] == 0) zeros++;
    if (n == 1) {
      puts("0");
      continue;
    }
    if (zeros == n) {
      puts("0");
      continue;
    }
    if (zeros == n - 1 && a[0] == 0) {
      puts("1");
      continue;
    }
    if (check(a, n, -1)) {
      puts("0");
      continue;
    }
    if (check(a, n, 0) || check(a, n, 1) || check(a, n, 2)) {
      puts("1");
      continue;
    }
    if (check2(a, n)) {
      puts("1");
      continue;
    }
    puts("2");
  }
  return 0;
}
