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 int INF = 1 << 28;
int N, vx[100020];
bool check(int skip) {
  if (N <= 1) return true;
  if (N <= 2 && skip >= 0) return true;
  int guilty = (skip == -2) ? 1 : 0;
  int c = INF, cr = INF, u = 0;
  for (int j = 0; j < (int)(N); ++j) {
    if (j == skip) continue;
    if (c == INF) {
      c = vx[j];
      if (c == 0) cr = 0, u = 0;
      continue;
    }
    if (cr == INF) {
      if (c == 0 && vx[j] != 0)
        guilty--;
      else
        u = cr = vx[j];
      continue;
    }
    if (c != 0 && ((u * cr) % c || abs(u * cr / c) > 10000)) {
      guilty--;
      continue;
    }
    int nex = (c) ? (u * cr / c) : 0;
    if (nex != vx[j])
      guilty--;
    else
      u = nex;
  }
  if (guilty < 0) return false;
  return true;
}
int main() {
  while (scanf("%d", &N) != EOF) {
    for (int j = 0; j < (int)(N); ++j) scanf("%d", vx + j);
    if (check(-1)) {
      puts("0");
      continue;
    }
    if (check(0)) {
      puts("1");
      continue;
    }
    if (check(1)) {
      puts("1");
      continue;
    }
    if (check(-2)) {
      puts("1");
      continue;
    }
    puts("2");
  }
}
