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[100005], b[1005], n;
int del(int i) {
  int j, k;
  k = 0;
  for (j = 0; j < n; j++) {
    if (j != i) b[k++] = a[j];
  }
  if (n == 3) {
    if (a[0] == 0 && a[1] != 0)
      return 2;
    else
      return 1;
  } else {
    for (j = 2; j < n - 1; j++) {
      if (b[j] * b[j - 2] != b[j - 1] * b[j - 1]) return 2;
    }
    return 1;
  }
}
int check() {
  int i, j, k = 0;
  int res = 9999;
  bool flag = true;
  for (i = 2; i < n; i++) {
    if (a[i] * a[i - 2] != a[i - 1] * a[i - 1]) {
      flag = false;
      if (i == n - 1)
        res = 1;
      else {
        res = min(res, del(i));
      }
      break;
    }
  }
  if (flag) return 0;
  res = min(res, del(0));
  res = min(res, del(1));
  return res;
}
int main() {
  int i, j, k, cnt0;
  while (scanf("%d", &n) != EOF) {
    cnt0 = 0;
    for (i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      if (a[i] == 0) cnt0++;
    }
    if (n == 1) {
      puts("0");
      continue;
    }
    if (n == 2) {
      if (a[0] == 0 && a[1] != 0)
        puts("1");
      else
        puts("0");
      continue;
    }
    if (cnt0 == n) {
      puts("0");
      continue;
    }
    if (cnt0 > 1) {
      if (a[0] != 0) {
        if (cnt0 == n - 2)
          puts("1");
        else if (cnt0 == n - 1)
          puts("0");
        else
          puts("2");
      } else {
        if (cnt0 == n - 1)
          puts("1");
        else
          puts("2");
      }
      continue;
    }
    int r1 = check();
    if (r1 == 0)
      puts("0");
    else if (r1 == 1)
      puts("1");
    else
      puts("2");
  }
  return 0;
}
