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;
bool bLeft[100000 + 5];
bool bRight[100000 + 5];
vector<int> v;
inline bool isNumber(int a, int b) {
  if (b == 0 && a != 0) return false;
  return true;
}
inline bool isSame(int a, int b, int c, int d) {
  if (b == 0 && a != 0) return false;
  if (d == 0 && c != 0) return false;
  if (a == 0 || c == 0) return a == 0 && c == 0;
  return a * d == b * c;
}
inline bool check() {
  for (int i = 1; i < v.size(); i++) {
    if (!isNumber(v[i], v[i - 1])) return false;
    if (i > 1) {
      if (!isSame(v[i], v[i - 1], v[i - 1], v[i - 2])) {
        return false;
      }
    }
  }
  return true;
}
int N;
int A[100000 + 5];
int main() {
  scanf("%d", &N);
  for (int i = 0; i < N; i++) {
    scanf("%d", &A[i]);
  }
  if (N <= 3) {
    v.clear();
    for (int i = 0; i < N; i++) v.push_back(A[i]);
    if (check())
      puts("0");
    else {
      bool flag = false;
      for (int i = 0; i < N; i++) {
        v.clear();
        for (int j = 0; j < N; j++) {
          if (i == j) continue;
          v.push_back(A[j]);
        }
        if (check()) {
          flag = true;
          break;
        }
      }
      if (flag)
        puts("1");
      else
        puts("0");
    }
  } else {
    for (int i = 1; i < N; i++) {
      if (i == 1) {
        bLeft[i] = true;
      } else {
        if (bLeft[i - 1] && isSame(A[i], A[i - 1], A[i - 1], A[i - 2])) {
          bLeft[i] = true;
        } else {
          bLeft[i] = false;
        }
      }
    }
    for (int i = N - 2; i >= 0; i--) {
      if (i == N - 2) {
        bRight[i] = true;
      } else {
        if (bRight[i + 1] && isSame(A[i + 1], A[i], A[i + 2], A[i + 1])) {
          bRight[i] = true;
        } else {
          bRight[i] = false;
        }
      }
    }
    if (bLeft[N - 1]) {
      puts("0");
    } else {
      bool flag = false;
      if (bRight[1]) {
        flag = true;
      }
      if (bRight[2] && isSame(A[2], A[0], A[3], A[2])) {
        flag = true;
      }
      for (int i = 2; i <= N - 3; i++) {
        if (bLeft[i - 1] && bRight[i + 1]) {
          if (isSame(A[i + 1], A[i - 1], A[i - 1], A[i - 2]) &&
              isSame(A[i + 2], A[i + 1], A[i + 1], A[i - 1])) {
            flag = true;
            break;
          }
        }
      }
      if (bLeft[N - 3] && isSame(A[N - 1], A[N - 3], A[N - 3], A[N - 4])) {
        flag = true;
      }
      if (bLeft[N - 2]) {
        flag = true;
      }
      if (flag)
        puts("1");
      else
        puts("2");
    }
  }
  return 0;
}
