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.lang.*;
import java.math.BigInteger;
import java.io.*;



public class Main {

	public static class Pair{
		int x;
		int y;
		public Pair(int a, int b) {
			x=a;
			y=b;
		}
		@Override
		public String toString() {
			return "Pair [x=" + x + ", y=" + y + "]";
		}
	}
	
	public static void main (String[] args) {
		FastReader sc=new FastReader();
		PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		int n=sc.nextInt();
		int[] tab=new int[n+1];
		tab[0]=0;
		for (int i=1;i<n+1;i++) {
			tab[i]=sc.nextInt();
		}
		int[] pat=new int[n];
		for (int i=0;i<n;i++) {
			pat[i]=tab[i+1]-tab[i];
		}

		int k=1;
		int tmp=0;
		
		ArrayList<Integer> ans=new ArrayList<>();
		while (k<=n) {
			boolean faux=true;
			ok :
			for (int i=0;i<k;i++) {
				tmp=i;
				while (tmp<n) {
					if (pat[i]!=pat[tmp]) {
						faux=false;
						break ok;
					}
					tmp+=k;
				}
			}
			if (faux==true) {
				ans.add(k);
			}
			k++;
		}
		out.println(ans.size());
		for (int i=0;i<ans.size();i++) {
			out.print(ans.get(i)+" ");
		}
		
		out.close();
	}
	
	static class FastReader
    {
        BufferedReader br;
        StringTokenizer st;
 
        public FastReader()
        {
            br = new BufferedReader(new InputStreamReader(System.in));
        }
        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException  e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
        int nextInt()
        {
            return Integer.parseInt(next());
        }
        long nextLong()
        {
            return Long.parseLong(next());
        }
        double nextDouble()
        {
            return Double.parseDouble(next());
        }
        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
	
}