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.*;

public class CP{
  public static OutputStream out=new BufferedOutputStream(System.out);
  static Scanner sc=new Scanner(System.in);
  static long mod=1000000007l;

  //nl-->neew line; //l-->line;  //arp-->array print;  //arpnl-->array print new line
  public static void nl(Object o) throws IOException{out.write((o+"\n").getBytes());}
  public static void l(Object o) throws IOException{out.write((o+"").getBytes());}
  public static void arp(int[] o) throws IOException{for(int i=0;i<o.length;i++) out.write((o[i]+" ").getBytes()); out.write(("\n").getBytes());}
  public static void arpnl(int[] o) throws IOException{for(int i=0;i<o.length;i++) out.write((o[i]+"\n").getBytes());}
  public static void scanl(long[] a,int n) {for(int i=0;i<n;i++) a[i]=sc.nextLong();}
  public static void scani(int[] a,int n) {for(int i=0;i<n;i++) a[i]=sc.nextInt();}
  public static void scan2D(int[][] a,int n,int m) {for(int i=0;i<n;i++) for(int j=0;j<m;j++) a[i][j]=sc.nextInt();}
  //
  static long cnt;
  static TreeSet<Integer> ans;
  static int n,m;
  static boolean[][] dp,vis;
  static HashMap<Pair,Boolean> arrl;


  public static void main(String[] args) throws IOException{
    long sttm=System.currentTimeMillis();
    long mod=1000000007l;

    int n=sc.nextInt();
    int[] a=new int[n+1];
    int[] dif=new int[n];
    for(int i=1;i<=n;i++){
        a[i]=sc.nextInt();
        dif[i-1]=a[i]-a[i-1];
    }
    ArrayList<Integer> ans=new ArrayList<Integer>();
    for(int i=0;i<n;i++){
        ArrayList<Integer> arrl=new ArrayList<Integer>();
        for(int k=0;k<=i;k++){
            arrl.add(dif[k]);
        }
        int l=arrl.size();
        boolean con=true;
        for(int j=i+1;j<n;j++){
            // nl((j-1-i)%l+" "+l+" "+i);
            if(arrl.get((j-i-1)%l)!=dif[j]){
                con=false;
                break;
            }
        }
        // nl(arrl);
        if(con) ans.add(i+1);
    }
    nl(ans.size());
    for(int elm:ans) l(elm+" "); nl("");

    out.flush();
  }
}

class Pair{
  int st,nd;
  Pair(int st,int nd){
    this.st=st;
    this.nd=nd;
  }
}
