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 val[100001], n;
int solve() {
  if (val[1] == 0) {
    int cnt = 0;
    for (int i = 2; i <= n; ++i)
      if (val[i] != 0) ++cnt;
    if (cnt <= 1) return cnt;
    for (int i = 2; i <= n; ++i)
      if (val[i] == 0) return 2;
    for (int i = 3; i <= n - 1; ++i)
      if (val[i] * val[i] != val[i - 1] * val[i + 1]) return 2;
    return 1;
  } else {
    int cnt = 0;
    for (int i = 2; i <= n; ++i)
      if (val[i] == 0) ++cnt;
    if (cnt == n - 1) return 0;
    if (cnt == n - 2) return 1;
    if (cnt < n - 2 && cnt > 1) return 2;
    int m = n, i = 1;
    if (cnt == 1) {
      for (n = 0; i <= m; ++i)
        if (val[i] != 0) val[++n] = val[i];
      for (int i = 2; i < n; ++i)
        if (val[i] * val[i] != val[i - 1] * val[i + 1]) return 2;
      return 1;
    }
    for (cnt = 0, i = 3; i <= n; ++i) {
      if (val[2] * val[i - 1] != val[i] * val[1]) ++cnt;
    }
    if (cnt == 0) return 0;
    for (cnt = 0, i = 3; i <= n; ++i) {
      if (val[3] * val[i - 1] != val[i] * val[2]) ++cnt;
    }
    if (cnt == 0) return 1;
    for (cnt = 0, i = 4; i <= n; ++i) {
      if (val[3] * val[i - 1] != val[i] * val[1]) ++cnt;
    }
    if (cnt == 0) return 1;
    int pos;
    for (cnt = 0, i = 3; i <= n; ++i) {
      if (val[2] * val[i - 1] != val[i] * val[1]) {
        ++cnt;
        pos = i;
        break;
      }
    }
    if (pos == n) return 1;
    if (val[2] * val[pos - 1] != val[pos + 1] * val[1]) return 2;
    for (i = pos + 2; i <= n; ++i) {
      if (val[2] * val[i - 1] != val[i] * val[1]) return 2;
    }
  }
  return 1;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", val + i);
  if (n == 1)
    printf("0\n");
  else {
    if (n == 2) {
      if (val[1] == 0 && val[2] != 0)
        printf("1\n");
      else
        printf("0\n");
      return 0;
    }
    if (n == 3) {
      if (val[1] == 0) {
        if (val[2] != 0 || val[3] != 0)
          printf("1\n");
        else
          printf("0\n");
      } else {
        printf("%d\n", val[2] * val[2] != val[1] * val[3]);
      }
      return 0;
    }
    printf("%d\n", solve());
  }
  return 0;
}
