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 a[110000];
int gcd(int a, int b) {
  if (!a || !b) return a + b;
  return gcd(b, a % b);
}
int gcds(int a, int b) { return gcd(abs(a), abs(b)); }
const int BAD = int(1e+9);
int n;
int check(int Ex) {
  int start, qch, qzn, cur;
  int can;
  if (Ex == 0) {
    start = 3;
    qch = a[2];
    qzn = a[1];
    cur = a[2];
    if (a[1] == 0 && a[2] != 0) return 2;
    can = 1;
  } else if (Ex == 1) {
    start = 3;
    qch = a[2];
    qzn = a[0];
    cur = a[2];
    if (a[0] == 0 && a[2] != 0) return 2;
    can = 1;
  } else {
    start = 2;
    qch = a[1];
    qzn = a[0];
    cur = a[1];
    if (a[0] == 0 && a[1] != 0) return 2;
    can = 0;
  }
  int g = gcds(qch, qzn);
  if (g != 0) {
    if (qch < 0) {
      qch *= -1;
      qzn *= -1;
    }
    qch /= g;
    if (qzn < 0) {
      qzn *= -1;
      qch *= -1;
    }
    qzn /= g;
  } else
    qch = qzn = 0;
  if (qzn == 0) {
    for (int i = start; i < n; i++) {
      if (a[i] != 0) can++;
    }
    return min(2, can);
  }
  for (int i = start; i < n; i++) {
    if (cur == BAD || abs(cur) % qzn != 0)
      cur = BAD;
    else {
      cur = cur / qzn * qch;
    }
    while (i < n && a[i] != cur) can++, i++;
  }
  return min(2, can);
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int ans = check(0);
  ans = min(ans, check(1));
  ans = min(ans, check(2));
  cout << ans << endl;
  return 0;
}
