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.io.*;
import java.util.Arrays;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.PrimitiveIterator.OfDouble;
import java.util.PriorityQueue;


public class BC {
static boolean ans=false;
static int a[];
static boolean [] vis=new boolean[5];
	public static void main(String[] args) throws IOException {
        Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
        int a[]=new int[n+1];
        a[0]=0;
        int b[]=new int[n];
        for (int i = 1; i < a.length; i++) {
			a[i]=sc.nextInt();
			b[i-1]=a[i-1]-a[i];
		}
        int k=1;
        int n1=0;
        PriorityQueue<Integer> pq=new PriorityQueue();
        for (k = 1; k <=b.length;k++ ) {
        	boolean f=true;	
		
        for (int i = 0; i < b.length; i++) {
			if(b[i]!=b[i%k]) {
				f=!f;
				break;
			}
		}
        if(f) {
        
        pq.add(k);
        }
        
        }
        System.out.println(pq.size());
        while(!pq.isEmpty()) {
        	System.out.print(pq.remove()+" ");
        }
	}
    private static void fillright(int n,int a[],int right[]) {
		Stack<Integer> st=new Stack<Integer>();
		for (int i = 0; i <n; i++) {
			while(!st.isEmpty()&&a[i]>=a[st.peek()]) {
				st.pop();
			}
			if(st.isEmpty()) {
				right[i]=-1;
			}
			else {
				right[i]=st.peek();
			}
			st.add(i);
		}
	}
	private static void swap(int[] a2, int i, int j) {
		int tmp=a2[i];
		a2[i]=a2[j];
		a2[j]=tmp;
	}
    
static class Scanner{
    StringTokenizer st;
    BufferedReader br;
    public Scanner(InputStream system){
        br=new BufferedReader(new InputStreamReader(system));
    }
    public Scanner (String file) throws IOException{
        br=new BufferedReader(new FileReader(file));
    }
    public String next() throws IOException{
        while (st==null || !st.hasMoreTokens())
            st=new StringTokenizer(br.readLine());
        return st.nextToken();
    }
    public  String nextLine() throws IOException{
        return br.readLine();

    }
    public int nextInt() throws IOException{
        return Integer.parseInt(next());

    } }



    
}
