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;
inline int eprintf(const char *format, ...) { return 0; }
bool eq(int a1, int a0, double d, bool e) {
  if (!a0) {
    if (a1) throw 0;
    return true;
  }
  if (!e) return true;
  return fabs((double)a1 / a0 - d) <= 1e-9;
}
int n;
void precalc(const vector<int> &a, vector<int> &e, vector<double> &ds,
             bool rev) {
  bool exd = false;
  e = vector<int>(n, -1);
  ds = vector<double>(n);
  double d;
  e[0] = 0;
  for (int i = 1; i < n; i++) {
    try {
      int a1 = a[i], a0 = a[i - 1];
      if (rev) swap(a1, a0);
      if (!eq(a1, a0, d, exd)) throw 0;
      if (!exd && a0) {
        d = (double)a1 / a0;
        exd = true;
      }
      e[i] = exd;
      ds[i] = d;
    } catch (...) {
      return;
    }
  }
}
int main() {
  while (scanf("%d", &n) >= 1) {
    vector<int> a(n);
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    vector<int> le, re;
    vector<double> ld, rd;
    precalc(a, le, ld, false);
    reverse(a.begin(), a.end());
    precalc(a, re, rd, true);
    reverse(a.begin(), a.end());
    reverse(re.begin(), re.end());
    reverse(rd.begin(), rd.end());
    if (le[n - 1] >= 0) {
      printf("0\n");
      goto end;
    }
    for (int sk = 0; sk < n; sk++) {
      try {
        bool e1 = false, e2 = false;
        double d1, d2;
        if (sk - 1 >= 0) {
          if (le[sk - 1] < 0) throw 0;
          e1 = le[sk - 1];
          d1 = ld[sk - 1];
        }
        if (sk + 1 < n) {
          if (re[sk + 1] < 0) throw 0;
          e2 = re[sk + 1];
          d2 = rd[sk + 1];
        }
        if (e1 && e2)
          if (fabs(d1 - d2) > 1e-9) continue;
        if (sk - 1 >= 0 && sk + 1 < n) {
          if (!eq(a[sk + 1], a[sk - 1], d1, e1)) continue;
          if (!eq(a[sk + 1], a[sk - 1], d2, e2)) continue;
        }
        printf("1\n");
        goto end;
      } catch (...) {
      }
    }
    printf("2\n");
  end:;
    break;
  }
  return 0;
}
