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 n, a[100005];
int main() {
  int i, cnt_zero = 0, lst, cur, nxt, ignore;
  bool flag = true;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 0) cnt_zero++;
  }
  if (n == 1 || cnt_zero == n || (n == 2 && cnt_zero == 0) ||
      (cnt_zero == n - 1 && a[1] != 0)) {
    printf("0\n");
    return 0;
  }
  if (cnt_zero == n - 1 || (cnt_zero == n - 2 && a[1] != 0)) {
    printf("1\n");
    return 0;
  }
  if (cnt_zero > 1) {
    printf("2\n");
    return 0;
  }
  if (cnt_zero == 1) {
    i = 1;
    while (i <= n && a[i] == 0) i++;
    lst = i, i++;
    while (i <= n && a[i] == 0) i++;
    cur = i, i++;
    while (i <= n && a[i] == 0) i++;
    nxt = i;
    while (nxt <= n) {
      if (a[cur] * a[cur] != a[nxt] * a[lst]) {
        printf("2\n");
        return 0;
      }
      lst = cur, cur = nxt;
      nxt++;
      while (nxt <= n && a[nxt] == 0) nxt++;
    }
    printf("1\n");
    return 0;
  }
  if (n == 3) {
    if (a[2] * a[2] == a[1] * a[3])
      printf("0\n");
    else
      printf("1\n");
    return 0;
  }
  for (i = 1; i <= n - 1; i++) {
    if (i == n - 1) {
      printf("0\n");
      return 0;
    }
    if (a[i + 1] * a[i + 1] != a[i] * a[i + 2]) break;
  }
  for (ignore = 1; ignore <= 2; ignore++) {
    i = 1;
    while (i <= n && i == ignore) i++;
    lst = i, i++;
    while (i <= n && i == ignore) i++;
    cur = i, i++;
    while (i <= n && i == ignore) i++;
    nxt = i;
    while (nxt <= n) {
      if (a[cur] * a[cur] != a[nxt] * a[lst]) break;
      lst = cur, cur = nxt, nxt++;
    }
    if (nxt > n) {
      printf("1\n");
      return 0;
    }
  }
  lst = 1, cur = 2, nxt = 3;
  while (nxt <= n) {
    if (a[cur] * a[cur] != a[nxt] * a[lst]) {
      if (flag) {
        nxt++;
        flag = false;
        continue;
      } else {
        printf("2\n");
        return 0;
      }
    }
    lst = cur, cur = nxt, nxt++;
  }
  printf("1\n");
  return 0;
}
