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 maxn = 100005;
const double epsi = 1e-8;
int n;
int a[maxn], c[maxn];
double b[maxn], d[maxn];
void init() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
}
int sign(double a) {
  if (a > epsi) return 1;
  if (a < -epsi) return -1;
  return 0;
}
bool del(int x) {
  int tot = 0;
  for (int i = 1; i <= n; i++)
    if (i != x) c[++tot] = a[i];
  for (int i = 2; i <= tot; i++) {
    d[i] = (double)c[i] / c[i - 1];
  }
  bool fz = true;
  for (int i = 3; i <= tot; i++)
    if (sign(d[2] - d[i]) != 0) {
      fz = false;
      break;
    }
  return fz;
}
void solve() {
  int cnt1, cnt0, cq1, cq2;
  double q1, q2;
  bool fz;
  if (n == 1) {
    printf("0\n");
    return;
  }
  if (n == 2) {
    if (a[1] == 0 && a[2] != 0)
      printf("1\n");
    else
      printf("0\n");
    return;
  }
  cnt1 = 0;
  cnt0 = 0;
  for (int i = 2; i <= n; i++) {
    if (a[i] != 0)
      cnt1++;
    else
      cnt0++;
  }
  if (a[1] == 0) {
    if (cnt1 == 0)
      printf("0\n");
    else if (cnt1 == 1)
      printf("1\n");
    else if (cnt0 == 0 && del(1))
      printf("1\n");
    else
      printf("2\n");
    return;
  }
  if (cnt0 > 0) {
    if (cnt0 > 1) {
      if (cnt1 == 0)
        printf("0\n");
      else if (cnt1 == 1)
        printf("1\n");
      else
        printf("2\n");
      return;
    } else {
      int tot = 0;
      for (int i = 1; i <= n; i++)
        if (a[i] != 0) c[++tot] = a[i];
      for (int i = 2; i <= tot; i++) {
        b[i] = (double)c[i] / c[i - 1];
      }
      q1 = b[2];
      fz = true;
      for (int i = 3; i <= tot; i++)
        if (sign(q1 - b[i]) != 0) {
          fz = false;
          break;
        }
      if (fz)
        printf("1\n");
      else
        printf("2\n");
      return;
    }
  }
  for (int i = 2; i <= n; i++) {
    b[i] = (double)a[i] / a[i - 1];
  }
  if (n == 3) {
    if (sign(b[2] - b[3]) == 0)
      printf("0\n");
    else
      printf("1\n");
    return;
  }
  fz = true;
  for (int i = 3; i <= n; i++)
    if (sign(b[2] - b[i]) != 0) {
      fz = false;
      break;
    }
  if (fz) {
    printf("0\n");
    return;
  }
  if (del(1)) {
    printf("1\n");
    return;
  }
  if (del(2)) {
    printf("1\n");
    return;
  }
  for (int i = 2; i <= n - 1; i++)
    if (sign(b[i + 1] - b[i]) != 0 &&
        (i == n - 1 || sign((double)a[i + 2] / a[i] - b[i]) == 0)) {
      if (del(i + 1)) {
        printf("1\n");
        return;
      }
    }
  printf("2\n");
}
int main() {
  init();
  solve();
  return 0;
}
