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.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n;
        int a;
        int a0 = 0;
        n = scanner.nextInt();

        int[] mas = new int[n];
        ArrayList<Integer> ans = new ArrayList<>(n);
        for (int i = 0; i < n; ++i) {
            a = scanner.nextInt();
            mas[i] = a - a0;
            a0 = a;
        }

        ArrayList<Integer> pi = new ArrayList<>(n);
        pi.add(0, 0);
        int j;
        int minh = 1, max = -1,MAX=-1;
        boolean bool = false;
        for (int i = 1; i < n; ++i) {
            j = pi.get(i - 1);
            while (j > 0 && (mas[i] != (mas[j])))
                j = pi.get(j - 1);
            if (mas[i] == (mas[j])) ++j;
            pi.add(i, j);
            if (j == 1) {
                minh = i;
            }
            if (j == 1 && !bool) {
                max = i;
                bool = true;
            }
            if(j>MAX)
                MAX=j;


        }

        int tmp=n - pi.get(n - 1);

//        System.out.println(max);
//        System.out.println(minh);
//        System.out.println(Arrays.toString(mas));
//        System.out.println(pi.toString());

        int temp=pi.get(n - 1);
        ans.add(n-temp);

        int i=0,m;
        int one=0,two=0;
//        if(MAX>temp) {

            for (int k = n - temp + 1; k < n; ++k) {
                // for (int i = 1; i < n; ++i)
                {
                    m = k;
                    i = 0;
                    do {
                        one = mas[m];
                        two = mas[i];
                        m++;
                        i++;
                    }while (one == two && m<n);
                    if (m == n && one == two)
                        ans.add(k);

                }
            }

            if (ans.get(ans.size() - 1) != n)
                ans.add(n);


        System.out.println(ans.size());
        Collections.sort(ans);
        for (Integer an : ans) System.out.print(an + " ");
    }
}