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.math.*;
import java.io.*;
public class main{
    public static void main(String args[] ) throws Exception { 
     Scanner sc = new Scanner(System.in);
     int n =sc.nextInt();
     int arr[]=new int[n+1];
     arr[0]=0;
     int temp = 0;
     for(int i=1;i<=n;i++) {
    	 arr[i] = sc.nextInt();
     }
     for(int i=n;i>=1;i--) {
    	 arr[i]=arr[i]-arr[i-1];
     }
    boolean dp[][] = new boolean[n+1][n+1];
    for(int j=1;j<=n;j++) {
    	dp[1][j] = (arr[j]==arr[1]);
    }
    for(int i=2;i<=n;i++) {
    	for(int j=i+1;j<=n;j++) {
    		if((j+i-1)>n) {
    			dp[i][j]=check(arr,j,n);
    			//System.out.println(i+" "+j+" "+dp[i][j]+" break");
    			break;
    		}
    		dp[i][j] = dp[i-1][j]&&(arr[i]==arr[j+i-1]);
    		//System.out.println(i+" "+j+" "+dp[i][j]);
    	}
    }
    Set<Integer> set  = new LinkedHashSet();
    for(int i=1;i<=n;i++) {
    	boolean val = true;
    	for(int j=i+1;j<=n;j+=i) {
    		if((j+i-1)>n) {
    			val&=check(arr,j,n);
    			break;
    		}
    		val&=dp[i][j];
    	}
    	if(val) {
    		set.add(i);
    	}
    }
    System.out.println(set.size());
    for(Integer i : set) {
    	System.out.print(i+" ");
    }
     
    }
    static boolean check(int[] arr,int i, int n) {
    	int j=1;
    	while(i<=n) {
    		if(arr[i]!=arr[j]) {
    			return false;
    		}
    		i++;
    		j++;
    	}
    	return true;
    }
}