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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.StreamTokenizer;

public class D {

    static StreamTokenizer st;
    static int n;
    public static void main(String[] args) throws IOException{
        st = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        n = nextInt();
        if (n==1) {
            System.out.println(0);
            return;
        }
        int[]a = new int[n+1];
        for (int i = 1; i <= n; i++) {
            a[i] = nextInt();
        }
        check(a);
//      for (int i = 1; i <= n/2; i++) {
//          int t = a[i];
//          a[i] = a[n-i+1];
//          a[n-i+1] = t;
//      }
//      check(a);
        System.out.println(2);
    }
    
    private static void check(int[] a) {
        int INF = (int) 1e9;
        int[]q = new int[n+1];
        boolean[]less_one = new boolean[n+1];
        for (int i = 1; i < n; i++) {
            if (a[i]==0) {
                if (a[i+1]==0)
                    q[i] = 0;
                else
                    q[i] = INF;
            }
            else {
                if (a[i+1] % a[i]==0)
                    q[i] = a[i+1]/a[i];
                else {
                    if (a[i] % a[i+1]==0) {
                        q[i] = a[i]/a[i+1];
                        less_one[i] = true;
                    }
                    else
                        q[i] = INF;
                }
            }
        }
        boolean f = true;
        for (int i = 1; i < n; i++) {
            if (q[i]==INF || q[i] != q[1] || less_one[i] != less_one[1]) {
                f = false;
                break;
            }
        }
        if (f) {
            System.out.println(0);
            System.exit(0);
        }
        f = true;
        for (int i = 2; i < n; i++) {
            if (q[i]==INF || q[i] != q[2] || less_one[i] != less_one[2]) {
                f = false;
                break;
            }
        }
        if (f || n==2) {
            System.out.println(1);
            System.exit(0);
        }
        boolean[]the_same2 = new boolean[n+1];
        if (q[n-1] != INF)
            the_same2[n-1] = true;
        for (int i = n-2; i >= 1; i--) {
            the_same2[i] = (the_same2[i+1] & (q[i]==q[i+1] && less_one[i]==less_one[i+1]) & (q[i] != INF));
        }
        boolean[]the_same1 = new boolean[n+1];
        if (q[1] != INF)
            the_same1[1] = true;
        for (int i = 2; i < n; i++) {
            the_same1[i] = (the_same1[i-1] & (q[i]==q[i-1] && less_one[i]==less_one[i-1]) & (q[i] != INF));
        }
        the_same1[0] = the_same2[n] = true;
        for (int i = 2; i <= n-1; i++) {
            int new_q = 0;
            boolean one_less = false;
            f = true;
            if (a[i-1]==0) {
                if (a[i+1]==0)
                    new_q = 0;
                else
                    f = false;
            }
            else {
                if (a[i+1] % a[i-1]==0)
                    new_q = a[i+1]/a[i-1];
                else {
                    if (a[i-1] % a[i+1]==0) {
                        new_q = a[i-1]/a[i+1];
                        one_less = true;
                    }
                    else
                        f = false;
                }
            }
            if (f && the_same1[i-2] && the_same2[i+1] && (new_q==q[i+1] && one_less==less_one[i+1]|| i+1==n) && (new_q==q[i-2] && one_less==less_one[i-2] || i-2==0)) {
                System.out.println(1);
                System.exit(0);
            }
        }
        f = true;
        for (int i = 1; i <= n-2; i++) {
            if (q[i]==INF || q[i] != q[1] || less_one[i] != less_one[1]) {
                f = false;
                break;
            }
        }
        if (f) {
            System.out.println(1);
            System.exit(0);
        }
    }

    private static int nextInt() throws IOException{
        st.nextToken();
        return (int) st.nval;
    }

}
