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.*;

public class Problem2 {
	

public static boolean test(int tab[],int n,int nx)
{
	int tabx[]=new int[nx];
	for (int i = 0; i < nx; i++) {
		tabx[i]=tab[i+1];
		//out.print( tabx[i]+ " ");
	}
	//out.println("");
	for (int i = 0; i < n; i++) {
		if(tabx[i%nx]!=tab[i+1])return false;
		//out.print("x "+i%nx+"="+tabx[i%nx] +" tab["+(i+1)+"]="+tab[i+1]+"|");
	}
	return true;
}
	public static void main(String[] args) {
	      MyScanner sc = new MyScanner();
	      out = new PrintWriter(new BufferedOutputStream(System.out));
	      int n=sc.nextInt();
	      int tab[]=new int[n+1];
	      int dtab[]=new int[n+1];

	      tab[0]=0;
	      int max=-1;
	      for (int i = 1; i <= n; i++) {
			tab[i]=sc.nextInt();
			//max=Math.max(tab[i], tab[i-1]);
		    dtab[i]=tab[i]-tab[i-1];  
			//out.println(dtab[i]);
		      

		}
	      
	      int counter=0;
	      LinkedList<Integer> result=new LinkedList<>();
	      for (int i = 1; i <= n; i++) {
	    	  if(test(dtab, n, i)) {
	    		  //out.print(i+" ");
	    		  result.add(i);
	    		  counter++;
	    	  }
		}
	      out.println(counter);
	      for (Integer integer : result) {
			out.print(integer+" ");
		}
	      out.close();
	   }

	     

	   //-----------PrintWriter for faster output---------------------------------
	   public static PrintWriter out;
	      
	   //-----------MyScanner class for faster input----------
	   public static class MyScanner {
	      BufferedReader br;
	      StringTokenizer st;
	 
	      public MyScanner() {
	         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;
	      }

	   }
	   //--------------------------------------------------------
	}


