Bajtek, known for his unusual gifts, recently got an integer array x_0, x_1, …, x_{k-1}.

Unfortunately, after a huge array-party with his extraordinary friends, he realized that he'd lost it. After hours spent on searching for a new toy, Bajtek found on the arrays producer's website another array a of length n + 1. As a formal description of a says, a_0 = 0 and for all other i (1 ≤ i ≤ n) a_i = x_{(i-1)mod k} + a_{i-1}, where p mod q denotes the remainder of division p by q.

For example, if the x = [1, 2, 3] and n = 5, then:

  * a_0 = 0, 
  * a_1 = x_{0mod 3}+a_0=x_0+0=1, 
  * a_2 = x_{1mod 3}+a_1=x_1+1=3, 
  * a_3 = x_{2mod 3}+a_2=x_2+3=6, 
  * a_4 = x_{3mod 3}+a_3=x_0+6=7, 
  * a_5 = x_{4mod 3}+a_4=x_1+7=9. 



So, if the x = [1, 2, 3] and n = 5, then a = [0, 1, 3, 6, 7, 9].

Now the boy hopes that he will be able to restore x from a! Knowing that 1 ≤ k ≤ n, help him and find all possible values of k — possible lengths of the lost array.

Input

The first line contains exactly one integer n (1 ≤ n ≤ 1000) — the length of the array a, excluding the element a_0.

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^6).

Note that a_0 is always 0 and is not given in the input.

Output

The first line of the output should contain one integer l denoting the number of correct lengths of the lost array.

The second line of the output should contain l integers — possible lengths of the lost array in increasing order.

Examples

Input

5
1 2 3 4 5


Output

5
1 2 3 4 5 

Input

5
1 3 5 6 8


Output

2
3 5 

Input

3
1 5 3


Output

1
3 

Note

In the first example, any k is suitable, since a is an arithmetic progression.

Possible arrays x:

  * [1]
  * [1, 1]
  * [1, 1, 1]
  * [1, 1, 1, 1] 
  * [1, 1, 1, 1, 1]



In the second example, Bajtek's array can have three or five elements.

Possible arrays x:

  * [1, 2, 2]
  * [1, 2, 2, 1, 2]



For example, k = 4 is bad, since it leads to 6 + x_0 = 8 and 0 + x_0 = 1, which is an obvious contradiction.

In the third example, only k = n is good.

Array [1, 4, -2] satisfies the requirements.

Note that x_i may be negative.
import java.util.*;
import java.io.*;
import java.math.BigInteger;

public class Main
{
    static long mod=1000000007;
    static long p=(long)1e10;
    public static void main(String[] args) throws Exception
    {
    	FastReader in=new FastReader();
    	PrintWriter pw=new PrintWriter(System.out);
    	int n=in.nextInt();
    	long[] arr=new long[n];
    	for(int i=0;i<n;i++)
    		arr[i]=in.nextInt();
    	int t=n,k=1;
    	ArrayList<Integer> al=new ArrayList<>();
    	while(t-->0)
    	{
    		long[] temp=new long[k];
    		boolean f=true;
    		temp[0]=arr[0];
    		for(int i=1;i<k;i++)
    			temp[i]=arr[i]-arr[i-1];
    		for(int i=k;i<n;i++)
    		{
    			if(arr[i-1]+temp[i%k]!=arr[i])
    			{
    				f=false;
    				break;
    			}
    		}
    		if(f)
    			al.add(k);
    		k++;
    	}
    	pw.println(al.size());
    	for(int i:al)
    		pw.print(i+" ");
    	pw.flush();
    }

}

class pair<A,B>
{
    A f;
    B s;

    public pair(A a,B b)
    {
        f=a;
        s=b;
    }
}

class FastReader
{
    BufferedReader br;
    StringTokenizer st;
 
    public FastReader()
    {
        br=new BufferedReader(new InputStreamReader(System.in));
    }
 
    public String next() throws IOException
    {
        if(st==null || !st.hasMoreElements())
        {
            st=new StringTokenizer(br.readLine());
        }
        return st.nextToken();
    }
 
    public int nextInt() throws IOException
    {
        return Integer.parseInt(next());
    }
 
    public long nextLong() throws IOException
    {
        return Long.parseLong(next());
    }
}