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];
int b[100005];
int n;
bool solve1(int m) {
  for (int i = 1; i < m - 1; i++)
    if (b[i] * 1LL * b[i] != b[i - 1] * 1LL * b[i + 1]) {
      return false;
    }
  return true;
}
void solve() {
  if (n <= 2 && a[0] != 0) {
    puts("0");
    return;
  }
  int nzc = 0;
  for (int i = 1; i < n; i++)
    if (a[i] != 0) ++nzc;
  if (!nzc) {
    puts("0");
    return;
  }
  if (nzc == 1) {
    puts("1");
    return;
  }
  int zc = 0;
  for (int i = 0; i < n; i++)
    if (a[i] == 0) ++zc;
  if (zc > 1) {
    puts("2");
    return;
  }
  if (zc == 1) {
    int m = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] != 0) {
        b[m++] = a[i];
      }
    }
    puts(solve1(m) ? "1" : "2");
    return;
  }
  int m = 0;
  for (int i = 0; i < n; i++) b[m++] = a[i];
  if (solve1(m)) {
    puts("0");
    return;
  }
  m = 0;
  for (int i = 1; i < n; i++) b[m++] = a[i];
  if (solve1(m)) {
    puts("1");
    return;
  }
  m = 0;
  for (int i = 0; i < n; i++)
    if (i != 1) b[m++] = a[i];
  if (solve1(m)) {
    puts("1");
    return;
  }
  m = 0;
  b[m++] = a[0];
  b[m++] = a[1];
  for (int i = 2; i < n; i++) {
    b[m++] = a[i];
    if (b[m - 2] * 1LL * b[m - 2] != b[m - 1] * 1LL * b[m - 3]) --m;
  }
  if (m >= n - 1)
    puts("1");
  else
    puts("2");
}
int main(void) {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  solve();
  return 0;
}
