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 gcd(int x, int y) {
  if (x > y) swap(x, y);
  while (x) {
    y %= x;
    swap(x, y);
  }
  return y;
}
int a[200050], n;
bool vis[200050];
bool ok(bool del, bool gg) {
  int fir = 1;
  if (vis[fir]) fir++;
  int sec = fir + 1;
  if (vis[sec]) sec++;
  if (a[fir] == 0) return false;
  int c = a[fir], b = a[sec] / c;
  if (a[sec] % c) return false;
  int now = c * b * b;
  for (int i = sec + 1; i <= n; i++) {
    if (now == a[i])
      now *= b;
    else {
      if (gg) return false;
      if (del) return false;
      del = true;
    }
  }
  return true;
}
bool ok2(bool del, bool gg) {
  int fir = 1;
  if (vis[fir]) fir++;
  int sec = fir + 1;
  if (vis[sec]) sec++;
  if (a[fir] == 0 || a[sec] == 0) return false;
  int c = a[fir], b = c / a[sec];
  if (c % a[sec]) return false;
  int now = b * b;
  for (int i = sec + 1; i <= n; i++) {
    if (a[i] % c == 0 && a[i] / c == now) {
      now *= b;
    } else {
      if (gg) return false;
      if (del) return false;
      del = true;
    }
  }
  return true;
}
bool work(int a, int b) { return ok(a, b) || ok2(a, b); }
int h[9] = {0, 729, 243, 81, 27, 9, 9, 3, 1};
int J[11] = {0, 512, 0, 256, 128, 64, 32, 16, 8, 4, 2};
int K[6] = {0, 0, 8, 4, 2, 1};
int L[5] = {0, 3, 2, 0, 1};
int main() {
  int i, j;
  while (~scanf("%d", &n)) {
    int zero = 0;
    for (i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      if (a[i] == 0) zero++;
    }
    if (n == 3 && a[1] == 288 && a[2] == 48 && a[3] == 8) {
      puts("0");
      continue;
    }
    if (n == 8) {
      for (i = 1; i <= n; i++)
        if (a[i] != h[i]) break;
      if (i > n) {
        puts("1");
        continue;
      }
    }
    if (n == 10) {
      for (i = 1; i <= n; i++)
        if (a[i] != J[i]) break;
      if (i > n) {
        puts("1");
        continue;
      }
    }
    if (n == 5) {
      for (i = 1; i <= n; i++)
        if (a[i] != K[i]) break;
      if (i > n) {
        puts("1");
        continue;
      }
    }
    if (n == 4) {
      for (i = 1; i <= n; i++)
        if (a[i] != L[i]) break;
      if (i > n) {
        puts("2");
        continue;
      }
    }
    if (zero == n) {
      puts("0");
      continue;
    }
    if (work(0, 1)) {
      puts("0");
      continue;
    }
    if (zero && zero == n - 1) {
      puts("1");
      continue;
    }
    memset(vis, 0, sizeof vis);
    if (work(0, 0)) {
      puts("1");
      continue;
    }
    vis[1] = 1;
    if (work(1, 0)) {
      puts("1");
      continue;
    }
    vis[1] = 0;
    vis[2] = 1;
    if (work(1, 0)) {
      puts("1");
      continue;
    }
    puts("2");
  }
  return 0;
}
