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[1 << 17], n;
bool eq(double &a, double &b) { return fabs(a - b) < 1e-9; }
bool checkZ() {
  int p, c;
  if (a[0] == 0) {
    c = 2;
    p = 1;
  } else if (a[1] == 0) {
    c = 2;
    p = 0;
  } else {
    c = 1;
    p = 0;
  }
  double d = 1.0 * a[c] / a[p];
  double t;
  while (c < n) {
    p = c;
    do {
      ++c;
    } while (c < n && a[c] == 0);
    if (c == n) break;
    t = 1.0 * a[c] / a[p];
    if (!eq(d, t)) return false;
  }
  return true;
}
int checkNZ() {
  int p = 0, c = 1;
  double d = 1.0 * a[c] / a[p];
  double t;
  int count = 0;
  while (c < n) {
    p = c;
    ++c;
    if (c == n) break;
    t = 1.0 * a[c] / a[p];
    if (!eq(d, t)) {
      if (count++) break;
      ++c;
      if (c == n) break;
      t = 1.0 * a[c] / a[p];
      if (!eq(d, t)) {
        ++count;
        break;
      }
    }
  }
  if (count < 2) {
    return count;
  }
  count = 1;
  c = 2;
  p = 1;
  d = 1.0 * a[c] / a[p];
  while (c < n) {
    p = c;
    ++c;
    if (c == n) break;
    t = 1.0 * a[c] / a[p];
    if (!eq(d, t)) {
      ++count;
      break;
    }
  }
  if (count < 2) {
    return count;
  }
  count = 1;
  c = 2;
  p = 0;
  d = 1.0 * a[c] / a[p];
  while (c < n) {
    p = c;
    ++c;
    if (c == n) break;
    t = 1.0 * a[c] / a[p];
    if (!eq(d, t)) {
      ++count;
      break;
    }
  }
  return count;
}
int main() {
  cin >> n;
  int i, nz = 0;
  for (i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    nz += (a[i] == 0);
  }
  if (n == 1) {
    printf("0\n");
    return 0;
  } else if (n == 2) {
    if (a[0] == 0) {
      if (a[1] == 0)
        printf("0\n");
      else
        printf("1\n");
    } else {
      printf("0\n");
    }
    return 0;
  } else if (nz == n) {
    printf("0\n");
    return 0;
  } else if (nz == n - 1) {
    if (a[0] == 0) {
      printf("1\n");
    } else {
      printf("0\n");
    }
    return 0;
  } else if (nz == n - 2) {
    if (a[0] != 0) {
      printf("1\n");
      return 0;
    }
  }
  if (nz == 1) {
    if (checkZ()) {
      printf("1\n");
      return 0;
    } else {
      printf("2\n");
      return 0;
    }
  } else if (nz == 0) {
    int res = checkNZ();
    printf("%d\n", res);
    return 0;
  } else {
    printf("2\n");
    return 0;
  }
  return 0;
}
