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 mm = 1e5 + 2;
int num[mm];
int n;
int main() {
  int i, j;
  int mul;
  int ans = 0;
  int s;
  while (scanf("%d", &n) != EOF) {
    ans = 0;
    s = 0;
    for (i = 0; i < n; i++) {
      scanf("%d", &num[i]);
      if (num[i] == 0) s++;
    }
    if (n == 1)
      printf("0\n");
    else if (n == 2) {
      if (num[0] == 0 && num[1] != 0)
        printf("1\n");
      else
        printf("0\n");
    } else if (s == n)
      printf("0\n");
    else if (s == n - 1) {
      if (num[0] != 0)
        printf("0\n");
      else
        printf("1\n");
    } else if (s >= 2 && s < n) {
      if (s == (n - 2) && num[0] != 0)
        printf("1\n");
      else
        printf("2\n");
    } else if (s == 1) {
      for (i = 0; i < n; i++)
        if (num[i] == 0) break;
      for (int k = i; k < n - 1; k++) num[k] = num[k + 1];
      mul = num[0] * num[n - 2];
      for (i = 1, j = n - 3; i <= j;) {
        if (num[i] * num[j] == mul) {
          i++;
          j--;
        } else {
          ans = 2;
          break;
        }
      }
      if (ans == 2)
        printf("2\n");
      else
        printf("1\n");
    } else if (n == 4) {
      if ((num[0] * num[2] == num[1] * num[1]) &&
          (num[1] * num[3] == num[2] * num[2]))
        printf("0\n");
      else if (num[0] * num[2] == num[1] * num[1])
        printf("1\n");
      else if (num[0] * num[3] == num[1] * num[1])
        printf("1\n");
      else if (num[0] * num[3] == num[2] * num[2])
        printf("1\n");
      else if (num[1] * num[3] == num[2] * num[2])
        printf("1\n");
      else
        printf("2\n");
    } else {
      if (num[0] * num[n - 1] == num[1] * num[n - 2]) {
        mul = num[0] * num[n - 1];
        i = 2;
        j = n - 3;
      } else if (num[1] * num[n - 1] == num[2] * num[n - 2]) {
        mul = num[1] * num[n - 1];
        ans++;
        i = 3;
        j = n - 3;
      } else if (num[0] * num[n - 1] == num[2] * num[n - 2]) {
        mul = num[0] * num[n - 1];
        ans++;
        i = 3;
        j = n - 3;
      } else if (num[0] * num[n - 2] == num[1] * num[n - 3]) {
        mul = num[1] * num[n - 3];
        ans++;
        i = 2;
        j = n - 4;
      } else if (num[0] * num[n - 1] == num[1] * num[n - 3]) {
        mul = num[0] * num[n - 1];
        ans++;
        i = 2;
        j = n - 4;
      } else
        ans = 2;
      if (ans >= 2)
        printf("2\n");
      else {
        for (; i <= j;) {
          if (num[i] * num[j] == mul) {
            if (i == j && (num[i] / num[i - 1] != num[i - 1] / num[i - 2]))
              ans++;
            else if (i == j - 1 &&
                     (num[j + 1] / num[j] != num[i - 1] / num[i - 2]))
              ans++;
            i++;
            j--;
          } else if (num[i + 1] * num[j] == mul) {
            ans++;
            i += 2;
            j--;
          } else if (num[i] * num[j - 1] == mul) {
            ans++;
            i++;
            j -= 2;
          } else {
            ans = 2;
            break;
          }
        }
        if (ans > 2) ans = 2;
        printf("%d\n", ans);
      }
    }
  }
  return 0;
}
