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 main() {
  int n, p, m = 2, i;
  double b, c;
  int a[100000];
  scanf("%d", &n);
  for (i = 0; i < n; i++) scanf("%d", &a[i]);
  if (n == 1 || (n == 2 && a[0] != 0)) {
    puts("0");
    return 0;
  }
  if (n == 2 && a[0] == 0) {
    if (a[1] == 0) {
      puts("0");
    } else {
      puts("1");
    }
    return 0;
  }
  if (a[0] == 0) {
    if (a[1] == 0) {
      p = 0;
      for (i = 0; i < n; i++) {
        if (a[i] != 0) p++;
      }
      if (p <= 1) {
        printf("%d\n", p);
        return 0;
      }
    }
    if (a[2] == 0) {
      p = 0;
      for (i = 0; i < n; i++) {
        if (a[i] != 0) p++;
      }
      if (p <= 1) {
        printf("%d\n", p);
        return 0;
      }
    }
  } else {
    c = a[0];
    b = (double)a[1] / a[0];
    p = 0;
    for (i = 1; i < n; i++) {
      if (fabs(a[i] - c * b) > 1e-6) {
        p++;
      } else {
        c = a[i];
      }
    }
    if (p < m) m = p;
    c = a[0];
    b = (double)a[2] / a[0];
    p = 1;
    for (i = 2; i < n; i++) {
      if (fabs(a[i] - c * b) > 1e-6) {
        p++;
      } else {
        c = a[i];
      }
    }
    if (p < m) m = p;
  }
  if (a[1] == 0) {
    if (a[2] == 0) {
      for (i = 3; i < n; i++) {
        if (a[i] != 0) break;
      }
      if (i == n) {
        if (m > 1) m = 1;
        printf("%d\n", m);
        return 0;
      }
    }
  } else {
    c = a[1];
    b = (double)a[2] / a[1];
    p = 1;
    for (i = 2; i < n; i++) {
      if (fabs(a[i] - c * b) > 1e-6) {
        p++;
      } else {
        c = a[i];
      }
    }
    if (p < m) m = p;
  }
  printf("%d\n", m);
  return 0;
}
