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
import java.io.*;

//Codeforces Beta Round #48 (Div. 1), D
public class GeometricalProblem {

	public static void main(String[] args) throws IOException {
		File file = new File("input.txt");
		if(file.exists())
			System.setIn(new FileInputStream(file));
		BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(r.readLine()), a[] = new int[n];
		String ss[] = r.readLine().split(" ");
		assert ss.length == n && n > 0;
		for(int i = 0; i < n; i++)
			a[i] = Integer.parseInt(ss[i]);

		System.out.println(solve(a, 0));
	}

	private static final double E = 1e-7;
	public static int solve(int a[], int start) {
		if(a.length - start == 1)
			return start;
		double b = a[start] != 0 ? (double)a[start + 1] / a[start] : 0;
		int result = start, prev = a[start] != 0 ? a[start + 1] : 0;
		for(int i = start + (a[start] != 0 ? 2 : 1); i < a.length; i++) {
			if(a[i] == prev * b || Math.abs(a[i] - prev * b) < E)
				prev = a[i];
			else {
				if(++result == 2)
					break;
			}
		}
		if(result == 2 && start == 0) {
			result = solve(a, 1);
			if(result == 2 && a[0] != 0) {
				a[1] = a[0];
				result = solve(a, 1);
			}
		}
		return result;
	}

}
