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 long double epsilon = 1e-9;
int main() {
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &v[i]);
  }
  bool tobe1 = false;
  if (v[0] == 0) {
    int hasNonZero = 0;
    for (int i = 1; i < n; i++)
      if (v[i] != 0) hasNonZero++;
    if (!hasNonZero) {
      printf("0\n");
      return 0;
    }
    if (v[1] == 0 || hasNonZero == 1) {
      if (hasNonZero == 1) {
        printf("1\n");
        return 0;
      } else {
        printf("2\n");
        return 0;
      }
    }
    vector<int> toTryWithout;
    toTryWithout.push_back(0);
    int prev, next, cur;
    int beg, end;
    bool is = false;
    for (int i = 0; i < 1; i++) {
      beg = 1;
      if (toTryWithout[i] == 0) beg = 2;
      end = n - 1;
      if (toTryWithout[i] == n - 1) end = n - 2;
      is = true;
      for (int j = beg; j < end; j++) {
        if (j == toTryWithout[i]) continue;
        if (toTryWithout[i] == j - 1)
          prev = j - 2;
        else
          prev = j - 1;
        if (toTryWithout[i] == j + 1)
          next = j + 2;
        else
          next = j + 1;
        if (v[j] * v[j] != v[prev] * v[next]) {
          is = false;
          break;
        }
        if (v[j] == 0 && ((v[next] == 0) ^ (v[prev] == 0))) {
          is = false;
          break;
        }
      }
      if (is) {
        printf("1\n");
        return 0;
      }
    }
    printf("2\n");
    return 0;
  } else {
    int hasNonZero = 0;
    for (int i = 1; i < n; i++)
      if (v[i] != 0) hasNonZero++;
    if (!hasNonZero) {
      printf("0\n");
      return 0;
    }
    if (hasNonZero == 1) tobe1 = true;
  }
  vector<int> toTryWithout;
  bool is = true;
  for (int i = 1; i < n - 1; i++) {
    if (v[i] * v[i] != v[i - 1] * v[i + 1]) {
      is = false;
      toTryWithout.push_back(i - 1);
      toTryWithout.push_back(i);
      toTryWithout.push_back(i + 1);
    }
    if (v[i] == 0 && ((v[i + 1] == 0) ^ (v[i - 1] == 0))) {
      is = false;
      toTryWithout.push_back(i - 1);
      toTryWithout.push_back(i);
      toTryWithout.push_back(i + 1);
    }
  }
  if (is) {
    printf("0\n");
    return 0;
  }
  int prev, next, cur;
  int beg, end;
  for (int i = 0; i < 3; i++) {
    beg = 1;
    if (toTryWithout[i] == 0) beg = 2;
    end = n - 1;
    if (toTryWithout[i] == n - 1) end = n - 2;
    is = true;
    for (int j = beg; j < end; j++) {
      if (j == toTryWithout[i]) continue;
      if (toTryWithout[i] == j - 1)
        prev = j - 2;
      else
        prev = j - 1;
      if (toTryWithout[i] == j + 1)
        next = j + 2;
      else
        next = j + 1;
      if (v[j] * v[j] != v[prev] * v[next]) {
        is = false;
        break;
      }
      if (v[j] == 0 && ((v[next] == 0) ^ (v[prev] == 0))) {
        is = false;
        break;
      }
    }
    if (is) {
      printf("1\n");
      return 0;
    }
  }
  if (tobe1) {
    printf("1\n");
    return 0;
  }
  printf("2\n");
  return 0;
}
