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;
int n;
vector<int> V;
vector<int> TMP;
bool is_geom() {
  int i, j, ch, zm;
  if (TMP.size() <= 1) return true;
  if (TMP[0] == 0) {
    for (j = 0; j < TMP.size(); ++j)
      if (TMP[j] != 0) return false;
    return true;
  }
  ch = TMP[1];
  zm = TMP[0];
  int cur = TMP[0], neo;
  for (j = 1; j < TMP.size(); ++j) {
    neo = (ch * cur) / zm;
    if (neo * zm != ch * cur) return false;
    if (neo != TMP[j]) return false;
    cur = neo;
  }
  return true;
}
int solve() {
  int i, j, ch, zm, br;
  TMP = V;
  if (is_geom()) return 0;
  TMP.clear();
  for (i = 0; i < n; ++i)
    if (i != 0) TMP.push_back(V[i]);
  if (is_geom()) return 1;
  TMP.clear();
  for (i = 0; i < n; ++i)
    if (i != 1) TMP.push_back(V[i]);
  if (is_geom()) return 1;
  br = 0;
  if (V[0] == 0) {
    for (j = 0; j < n; ++j)
      if (V[j] != 0) ++br;
    if (br > 1) return 2;
    return 1;
  }
  ch = V[1];
  zm = V[0];
  int cur = V[0], neo;
  for (j = 1; j < V.size(); ++j) {
    neo = (ch * cur) / zm;
    if (neo * zm != ch * cur) {
      ++br;
      continue;
    }
    if (neo != V[j]) {
      ++br;
      continue;
    }
    cur = neo;
  }
  if (br < 2) return 1;
  return 2;
}
int main() {
  int i, j, k, l;
  scanf("%d", &n);
  for (i = 0; i < n; ++i) {
    scanf("%d", &l);
    V.push_back(l);
  }
  TMP = V;
  printf("%d\n", solve());
  return 0;
}
