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 M = 1100000;
int a[M];
const double eps = 1e-4;
int main() {
  int n;
  double b, c;
  while (scanf("%d", &n) == 1) {
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    if (n <= 2) {
      if (n == 2 & a[1] == 0 && a[2] != 0)
        printf("1\n");
      else
        printf("0\n");
      continue;
    }
    c = a[1];
    if (a[1] != 0)
      b = (double)a[2] / c;
    else
      b = 0;
    bool flag = true;
    for (int i = 3; i <= n; i++) {
      if (fabs(a[i] - a[i - 1] * b) < eps)
        continue;
      else
        flag = false;
    }
    if (a[1] == 0 && a[2] != 0) flag = false;
    if (flag) {
      printf("%d\n", 0);
      continue;
    }
    c = a[2];
    if (a[2] != 0)
      b = (double)a[3] / c;
    else
      b = 0;
    flag = true;
    for (int i = 4; i <= n; i++) {
      if (fabs(a[i] - a[i - 1] * b) < eps)
        continue;
      else
        flag = false;
    }
    if (a[2] == 0 && a[3] != 0) flag = false;
    if (flag) {
      printf("%d\n", 1);
      continue;
    }
    c = a[1];
    if (a[1] == 0)
      b = 0;
    else
      b = (double)a[3] / c;
    flag = true;
    for (int i = 4; i <= n; i++) {
      if (fabs(a[i] - a[i - 1] * b) < eps)
        continue;
      else
        flag = false;
    }
    if (a[1] == 0 && a[3] != 0) flag = false;
    if (flag) {
      printf("%d\n", 1);
      continue;
    }
    c = a[1];
    if (a[1] == 0)
      b = 0;
    else
      b = (double)a[2] / c;
    if (a[1] == 0 && a[2] != 0) {
      printf("%d\n", 2);
      continue;
    }
    int cnt = 0;
    int st[10];
    for (int i = 3; i <= n; i++) {
      if (fabs(a[i] - a[i - 1] * b) < eps)
        continue;
      else {
        cnt++;
        if (cnt > 2) break;
        st[cnt - 1] = i;
      }
    }
    if (cnt > 2)
      printf("%d\n", 2);
    else {
      if (cnt == 1) {
        if (st[0] == n || fabs(a[st[0] + 1] - a[st[0] - 1] * b) < eps ||
            fabs(a[st[0]] - a[st[0] - 2] * b) < eps)
          printf("%d\n", 1);
        else
          printf("%d\n", 2);
      } else {
        if (st[1] - st[0] > 1)
          printf("%d\n", 2);
        else {
          if (fabs(a[st[1]] - a[st[0] - 1] * b) < eps)
            printf("%d\n", 1);
          else
            printf("%d\n", 2);
        }
      }
    }
  }
  return 0;
}
