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 double eps = 10e-6;
int main() {
  int n;
  scanf("%d", &n);
  vector<int> a(n);
  int null = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 0) {
      null++;
    }
  }
  if (null == n || (null == n - 1 && a[0] != 0) || n == 1) {
    printf("0");
    return 0;
  }
  if (null > 0 && null == n - 2 && a[0] != 0) {
    printf("1");
    return 0;
  }
  if (null == n - 1) {
    printf("1");
    return 0;
  }
  if (null > 1) {
    printf("2");
    return 0;
  }
  if (n == 2) {
    if (null == 0) {
      printf("0");
    } else {
      printf("1");
    }
    return 0;
  }
  vector<double> b;
  if (null == 1) {
    int last = 0;
    if (a[0] == 0) {
      last = 1;
    }
    for (int i = last + 1; i < n; i++) {
      if (a[i] == 0) {
        continue;
      }
      b.push_back(1.0 * a[i] / a[last]);
      last = i;
    }
    for (size_t i = 1; i < b.size(); i++) {
      if (b[i] != b[i - 1]) {
        printf("2");
        return 0;
      }
    }
    printf("1");
    return 0;
  }
  for (int i = 1; i < n; i++) {
    b.push_back(1.0 * a[i] / a[i - 1]);
  }
  int x = 0;
  vector<int> fa;
  vector<int> m;
  for (size_t i = 1; i < b.size(); i++) {
    if (b[i] != b[i - 1]) {
      x++;
      m.push_back(i);
    } else {
      fa.push_back(i);
    }
  }
  sort(fa.begin(), fa.end());
  if (n == 3) {
    if (x == 1) {
      printf("1");
    } else {
      printf("0");
    }
    return 0;
  }
  if (n == 4) {
    if (x == 0) {
      printf("0");
    } else {
      if (abs(b[0] * b[1] / b[2] - 1) <= eps ||
          abs(b[1] * b[2] / b[0] - 1) <= eps || b[0] == b[1] || b[1] == b[2]) {
        printf("1");
      } else {
        printf("2");
      }
    }
    return 0;
  }
  if (x == 0) {
    printf("0");
  }
  if (x == 1) {
    printf("1");
  }
  if (x > 3) {
    printf("2");
  }
  if (x == 3 || x == 2) {
    int f;
    if (fa.size() > 1) {
      f = fa[1];
    } else {
      f = fa[0];
    }
    if (m[0] == 1 && m[1] == 2 && abs(b[0] * b[1] / b[f] - 1) <= eps) {
      printf("1");
      return 0;
    }
    if (x == 3 && m[2] - m[1] >= 3) {
      printf("2");
      return 0;
    }
    if (m[1] - m[0] < 3 && abs(b[m[0]] * b[m[0] + 1] / b[f] - 1) <= eps) {
      printf("1");
    } else {
      printf("2");
    }
  }
  return 0;
}
