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;
const int maxn = 100005;
int a[maxn];
int n;
bool ok(int t1, int t2, int t3) {
  if (t2 < 0 || t2 >= n) return true;
  if (t1 < 0 && t3 >= n) return true;
  if (t1 < 0) {
    if (a[t2] == 0 && a[t3] != 0) return false;
    return true;
  }
  if (t3 >= n) {
    if (a[t1] == 0 && a[t2] != 0) return false;
    return true;
  }
  if (a[t1] * a[t3] != a[t2] * a[t2]) return false;
  if (a[t2] != 0) return true;
  return a[t3] == 0;
}
bool vis[maxn], tm1[maxn], tm2[maxn];
int main() {
  while (scanf("%d", &n) != EOF) {
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
    }
    bool over = false;
    for (int i = 0; i < n; i++) {
      vis[i] = ok(i - 1, i, i + 1);
      if (!vis[i]) over = true;
    }
    if (!over) {
      printf("0\n");
      continue;
    }
    for (int i = 0; i < n; i++) {
      tm1[i] = vis[i];
      if (i != 0) tm1[i] = (tm1[i] && tm1[i - 1]);
    }
    for (int i = n - 1; i >= 0; i--) {
      tm2[i] = vis[i];
      if (i != n - 1) tm2[i] = (tm2[i] && tm2[i + 1]);
    }
    over = false;
    for (int i = 0; i < n; i++) {
      if (i >= 2 && !tm1[i - 2]) continue;
      if (i + 2 < n && !tm2[i + 2]) continue;
      if (!ok(i - 2, i - 1, i + 1)) continue;
      if (!ok(i - 1, i + 1, i + 2)) continue;
      over = true;
      break;
    }
    printf("%d\n", over ? 1 : 2);
  }
  return 0;
}
