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 INF = 1 << 30;
int n, in[100010], cnt, dyn[100010];
void check() {
  if (cnt < 2) {
    printf("%d\n", cnt);
    exit(0);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &in[i]);
  for (int i = n - 1; i >= 0; i--) dyn[i] = dyn[i + 1] + (in[i] == 0 ? 1 : 0);
  if (n == 2) {
    if (in[0] == 0 && in[1] != 0)
      printf("1\n");
    else
      printf("0\n");
    return 0;
  }
  if (dyn[0] > 1) {
    int ans = INF;
    if (dyn[0] == n) ans = min(ans, 0);
    if (dyn[1] >= n - 2) ans = min(ans, n - 1 - dyn[1]);
    if (dyn[2] == n - 2) ans = min(ans, n - 2);
    if (ans < 2)
      printf("%d\n", ans);
    else
      printf("2\n");
    return 0;
  }
  bool t = true;
  for (int i = 2; i < n; i++)
    if (in[i] * in[i - 2] != in[i - 1] * in[i - 1]) t = false;
  if (t) {
    printf("0\n");
    return 0;
  }
  if (n == 3) {
    if (in[1] == 0) goto l;
  }
  t = true;
  for (int i = 2; i < n; i++)
    if (in[i] * in[1] != in[2] * in[i - 1]) t = false;
  if (t) {
    printf("1\n");
    return 0;
  }
l:;
  cnt = 0;
  for (int i = 1; i < n; i++) {
    if (in[i] * in[0] != in[1] * in[i - 1]) {
      cnt++;
      if (i < n - 1 && in[i + 1] * in[0] != in[1] * in[i - 1]) cnt++;
      i++;
    }
  }
  check();
  cnt = 0;
  for (int i = 1; i < n; i++) {
    if (in[i] * in[1] != in[2] * in[i - 1]) {
      cnt++;
      if (i < n - 1 && in[i + 1] * in[1] != in[2] * in[i - 1]) cnt++;
      i++;
    }
  }
  check();
  cnt = 0;
  for (int i = 1; i < n; i++) {
    if (in[i] * in[0] != in[2] * in[i - 1]) {
      cnt++;
      if (i < n - 1 && in[i + 1] * in[0] != in[2] * in[i - 1]) cnt++;
      i++;
    }
  }
  check();
  printf("2\n");
  return 0;
}
