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 seq[100010], seq1[100010], seq2[100010], seq3[100010], superfinalmark = 0;
int main() {
  int n, nzero = 0, finalmark = -1, it1 = 0, it2 = 0, it3 = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &seq[i]);
    if (seq[i] == 0) nzero++;
  }
  if (nzero == n) {
    printf("0\n");
    return 0;
  }
  if (nzero == n - 1 && seq[0] != 0) {
    printf("0\n");
    return 0;
  }
  if (nzero == n - 1 && n > 2) {
    printf("1\n");
    return 0;
  }
  if (nzero == n - 2 && seq[0] != 0 && n >= 4) {
    printf("1\n");
    return 0;
  }
  if (n == 1) {
    printf("0\n");
    return 0;
  }
  if (n == 2) {
    if (nzero == 0) {
      printf("0\n");
      return 0;
    } else if (seq[0] > 0) {
      printf("0\n");
      return 0;
    } else {
      printf("1\n");
      return 0;
    }
  }
  if (n == 3 && nzero == 2 && seq[0] != 0) {
    printf("1\n");
    return 0;
  }
  if (nzero < n && nzero >= 2) {
    printf("2\n");
    return 0;
  }
  if (n == 3)
    if (nzero == 1) {
      printf("1\n");
      return 0;
    }
  if (nzero == 1) {
    int p;
    for (p = 0; seq[p] != 0; p++)
      ;
    for (int i = p + 1; i < n; i++) seq[i - 1] = seq[i];
    n--;
  }
  for (int i = 0; i < n - 2; i++)
    if (((double)seq[i + 1] / seq[i] - (double)seq[i + 2] / seq[i + 1]) >
            1e-9 ||
        ((double)seq[i + 1] / seq[i] - (double)seq[i + 2] / seq[i + 1]) <
            -1e-9) {
      if (nzero == 1) {
        printf("2\n");
        return 0;
      } else {
        finalmark = i;
        break;
      }
    }
  if (finalmark == -1) {
    printf("%d\n", nzero);
    return 0;
  }
  if (n == 3) {
    printf("1\n");
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (i != finalmark) seq1[it1++] = seq[i];
    if (i != finalmark + 1) seq2[it2++] = seq[i];
    if (i != finalmark + 2) seq3[it3++] = seq[i];
  }
  for (int i = 0; i < n - 3; i++)
    if (((double)seq1[i + 1] / seq1[i] - (double)seq1[i + 2] / seq1[i + 1]) >
            1e-9 ||
        ((double)seq1[i + 1] / seq1[i] - (double)seq1[i + 2] / seq1[i + 1]) <
            -1e-9) {
      superfinalmark++;
      break;
    }
  for (int i = 0; i < n - 3; i++)
    if (((double)seq2[i + 1] / seq2[i] - (double)seq2[i + 2] / seq2[i + 1]) >
            1e-9 ||
        ((double)seq2[i + 1] / seq2[i] - (double)seq2[i + 2] / seq2[i + 1]) <
            -1e-9) {
      superfinalmark++;
      break;
    }
  for (int i = 0; i < n - 3; i++)
    if (((double)seq3[i + 1] / seq3[i] - (double)seq3[i + 2] / seq3[i + 1]) >
            1e-9 ||
        ((double)seq3[i + 1] / seq3[i] - (double)seq3[i + 2] / seq3[i + 1]) <
            -1e-9) {
      superfinalmark++;
      break;
    }
  if (superfinalmark < 3)
    printf("1\n");
  else
    printf("2\n");
  return 0;
}
