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 a[100100];
inline bool cmp1(int a, int b) {
  if (a < 0) a = -a;
  if (b < 0) b = -b;
  return a < b;
}
int n;
inline bool check(int x) {
  if (x < 0) {
    if (-x > 10000) return false;
  } else {
    if (x > 10000) return false;
  }
  return true;
}
int mpow(int x, int b) {
  if (b == 0) return 1;
  if (!check(x)) return 100000007;
  if (b & 1) {
    int pr = mpow(x, b - 1);
    if (pr == 100000007) return 100000007;
    pr *= x;
    if (!check(pr)) return 100000007;
    return pr;
  } else {
    int pr = mpow(x * x, b >> 1);
    if (!check(pr)) return 100000007;
    return pr;
  }
}
inline bool comply(int f, int c, int b, bool canfail) {
  int pr;
  int skip = 0;
  for (int i = f; i < n; i++)
    if (c * (pr = mpow(b, i - f - skip)) != a[i] and pr != 100000007) {
      if (canfail)
        canfail = false, skip = 1;
      else
        return false;
    }
  return true;
}
int gcd(int a, int b) {
  if (!b)
    return a;
  else
    return gcd(b, a % b);
}
int getb(int a, int b) {
  int r = 1;
  if ((a < 0 and b > 0) or (a > 0 and b < 0)) {
    r = -1;
    if (a < 0) a = -a;
    if (b < 0) b = -b;
  }
  return gcd(a, b) == 0 ? 0 : r * (b / gcd(a, b));
}
int doit() {
  if (n == 1) {
    return 0;
  }
  if (comply(0, a[0], getb(a[0], a[1]), false) and getb(a[0], a[1]) != 0) {
    return 0;
  }
  if (comply(0, a[0], getb(a[0], a[1]), true) and getb(a[0], a[1]) != 0) {
    return 1;
  }
  if (n >= 3 and comply(0, a[0], getb(a[0], a[2]), true) and
      getb(a[0], a[1]) != 0) {
    return 1;
  }
  if (n >= 3 and comply(1, a[1], getb(a[1], a[2]), false) and
      getb(a[0], a[1]) != 0) {
    return 1;
  }
  return 2;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  int nz = 0;
  for (int i = 1; i < n; i++)
    if (a[i] != 0) nz++;
  if (nz == 0) {
    cout << 0 << endl;
    return 0;
  }
  int mx = 2;
  if (nz == 1) {
    mx = 1;
  }
  int s1 = doit();
  for (int i = 0; i < n / 2; i++) swap(a[i], a[n - i - 1]);
  int s2 = doit();
  cout << min(mx, min(s1, s2)) << endl;
}
