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 << 30;
int NOD(int x, int y) {
  if (x < 0) x *= -1;
  if (y < 0) y *= -1;
  while (x && y)
    if (x > y)
      x %= y;
    else
      y %= x;
  return x + y;
}
int NOK(int x, int y) { return x / NOD(x, y) * y; }
struct drob {
  int chis;
  int znam;
  drob(int C = 1, int Z = 1) {
    if (C < 0) C *= -1, Z *= -1;
    chis = C;
    znam = Z;
    int div = NOD(chis, znam);
    chis /= div;
    znam /= div;
  }
};
bool operator<(const drob& a, const drob& b) {
  return (a.chis * b.znam < b.chis * a.znam);
}
bool operator==(const drob& a, const drob& b) {
  return make_pair(a.chis, a.znam) == make_pair(b.chis, b.znam);
}
bool Try(vector<int> a, int x) {
  map<drob, int> M;
  M.clear();
  a.erase(a.begin() + x);
  for (int i = 0; i < a.size() - 1; i++) M[drob(a[i + 1], a[i])]++;
  return M.size() <= 1;
}
map<drob, int> M;
int n;
vector<int> a;
int kolofnul = 0;
int nulpos = -1;
int perm[4][3] = {{0, 1, 2}, {0, 2, 3}, {1, 2, 3}, {0, 1, 3}};
int main() {
  cin >> n;
  a.resize(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (!a[i]) kolofnul++, nulpos = i;
  }
  if (n == 1) {
    cout << 0;
    return 0;
  }
  if (kolofnul) {
    if (kolofnul == n) {
      cout << 0;
    } else if (kolofnul == n - 1) {
      if (a[0] != 0)
        cout << 0;
      else
        cout << 1;
    } else if (kolofnul == 1) {
      if (Try(a, nulpos))
        cout << 1;
      else
        cout << 2;
    } else if (kolofnul == n - 2 && a[0] != 0) {
      cout << 1;
    } else
      cout << 2;
    return 0;
  }
  if (n == 2) {
    cout << 0;
    return 0;
  }
  if (n == 3) {
    if (drob(a[2], a[1]) == drob(a[1], a[0]))
      cout << 0;
    else
      cout << 1;
    return 0;
  }
  if (n == 4) {
    for (int i = 0; i < n - 1; i++) M[drob(a[i + 1], a[i])]++;
    if (M.size() == 1) {
      cout << 0;
      return 0;
    } else {
      for (int i = 0; i < 4; i++) {
        int x = a[perm[i][0]];
        int y = a[perm[i][1]];
        int z = a[perm[i][2]];
        if (drob(y, x) == drob(z, y)) {
          cout << 1;
          return 0;
        }
      }
      cout << 2;
    }
    return 0;
  }
  for (int i = 0; i < n - 1; i++) M[drob(a[i + 1], a[i])]++;
  if (M.size() > 3)
    cout << 2;
  else if (M.size() == 1)
    cout << 0;
  else {
    if (n <= 700) {
      for (int i = 0; i < n; i++)
        if (Try(a, i)) {
          cout << 1;
          return 0;
        }
      cout << 2;
      return 0;
    }
    drob best = M.begin()->first;
    map<drob, int>::iterator it;
    for (it = M.begin(); it != M.end(); it++)
      if (it->second > M[best]) best = it->first;
    if (M[best] < (n - 1 - 2)) {
      cout << 2;
      return 0;
    }
    for (int i = 0; i < n; i++) {
      drob prev(inf, 1);
      drob next(inf, 1);
      if (i != 0) prev = drob(a[i], a[i - 1]);
      if (i != n - 1) next = drob(a[i + 1], a[i]);
      if (!(prev == best) || !(next == best)) {
        if (Try(a, i)) {
          cout << 1;
          return 0;
        }
      }
    }
    cout << 2;
    return 0;
  }
}
