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 f[100005], n;
int gcd(int a, int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
bool check(int st, int a, int b) {
  int p;
  if (a == 0) {
    b = 0;
    a = 1;
  }
  int now = f[st];
  for (int i = st + 1; i < n; i++) {
    if (gcd(now * b, a) != a) return 0;
    now = now * b / a;
    if (f[i] != now) return 0;
  }
  return 1;
}
bool check2(int st, int a, int b) {
  int p;
  if (a == 0) {
    b = 0;
    a = 1;
  }
  int now = f[st], cnt = 0;
  for (int i = st + 1; i < n; i++) {
    if (gcd(now * b, a) != a) return 0;
    now = now * b / a;
    while (f[i] != now) {
      cnt++;
      i++;
      if (cnt == 2) return 0;
      if (i == n) return 1;
    }
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", f + i);
  if (check(0, f[0], f[1]))
    puts("0");
  else if (check(1, f[1], f[2])) {
    puts("1");
  } else if (check(2, f[0], f[2])) {
    puts("1");
  } else if (check2(0, f[0], f[1])) {
    puts("1");
  } else
    puts("2");
}
