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;
template <class T>
int size(const T& a) {
  return (int)a.size();
}
template <class T>
T sqr(const T& a) {
  return a * a;
}
int gcd(int a, int b) {
  if (a < 0) a = -a;
  if (b < 0) b = -b;
  while (a && b) (a > b ? a %= b : b %= a);
  return a + b;
}
struct Q {
  int p, q;
  Q() {}
  Q(int a, int b) {
    p = a;
    q = b;
    if (p == 0) {
      q = 1;
    }
    if (q < 0) {
      p = -p;
      q = -q;
    }
    int d = gcd(p, q);
    p /= d;
    q /= d;
  }
  bool operator<(const Q& z) const { return p < z.p || p == z.p && q < z.q; }
  bool operator==(const Q& z) const { return p == z.p && q == z.q; }
};
int a[100100];
int getr(const Q& b, int n) {
  if (b.q == 0) return 2;
  int nd = 0;
  for (int i = 1, j = 0; i < n; i++) {
    Q q(a[i], a[j]);
    while (j + 1 < i && !(q == b)) {
      j++;
    }
    if (Q(a[i], a[j]) == b) {
      j = i;
    } else {
      nd++;
    }
  }
  return nd;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  if (n == 1) {
    printf("0");
    return 0;
  }
  map<Q, int> q;
  for (int i = 1; i < n; i++) {
    q[Q(a[i], a[i - 1])]++;
  }
  Q b;
  int br = -1, nd = 2;
  if (n <= 3) nd = 1;
  for (map<Q, int>::iterator it = q.begin(); it != q.end(); it++) {
    if (it->second > br) {
      br = it->second;
      b = it->first;
    }
  }
  if (n < 10) {
    for (map<Q, int>::iterator it = q.begin(); it != q.end(); it++) {
      nd = min(nd, getr(it->first, n));
    }
  } else {
    nd = min(nd, getr(b, n));
  }
  printf("%d", min(nd, 2));
  return 0;
}
