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.
//package math_codet;

import java.io.*;
import java.util.*;
 /******************************************
*    AUTHOR:         AMAN KUMAR SINGH        *
*    INSTITUITION:   KALYANI GOVERNMENT ENGINEERING COLLEGE  *
******************************************/
public class lets_do {
    InputReader in;
    PrintWriter out;
    Helper_class h;
    final long mod = 1000000007;
    public static void main(String[] args) throws java.lang.Exception{
        new lets_do().run();
    }
    void run() throws Exception{
        in=new InputReader();
        out = new PrintWriter(System.out);
        h = new Helper_class();
        int t = 1;
        while(t-- > 0)
            solve();
        out.flush();    
        out.close();
    }
    TreeMap<Integer, Integer> tmap = new TreeMap<Integer, Integer>();
    void solve(){
        int n = h.ni();
        int[] arr = new int[n + 1];
        int i = 0;
        for(i = 1; i <= n; i++)
            arr[i] = h.ni();
        int cnt = 0;
        StringBuffer str = new StringBuffer();
        for(int k = 1; k <= n; k++){
            int[] x = new int[k];
            x[0] = arr[1];
            for(i = 1; i < k && i < n; i++)
                x[i] = arr[i + 1] - arr[i];
            //h.pn(Arrays.toString(x));
            boolean ok = true;
            for(i = k; i < n; i++){
                if(arr[i + 1] != x[i % k] + arr[i])
                    ok = false;
            }
            if(ok) {
                cnt++;
                str.append(k+" ");
            }
        }
        h.pn(cnt);
        h.pn(str);
    }
    void add(int x){
        Integer cnt = tmap.get(x);
        if(cnt == null)
            tmap.put(x, 1);
        else
            tmap.put(x, ++cnt);
    }
    void remove(int x){
        Integer cnt = tmap.get(x);
        if(cnt == 1)
            tmap.remove(x);
        else
            tmap.put(x, --cnt);
    }
    final Comparator<Entity> com = new Comparator<Entity>(){
        public int compare(Entity x, Entity y){
            int xx = Integer.compare(x.a, y.a);
            if(xx == 0)
                return Integer.compare(x.b, y.b);
            else
                return xx;
        }
    };
    class Entity{
        int a;
        int b;
        Entity(int p, int q){
            a = p;
            b = q;
        }
    }
    class Helper_class{
        long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
        int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
        int bitcount(long n){return (n==0)?0:(1+bitcount(n&(n-1)));}
        void p(Object o){out.print(o);}
        void pn(Object o){out.println(o);}
        void pni(Object o){out.println(o);out.flush();}
        String n(){return in.next();}
        String nln(){return in.nextLine();}
        int ni(){return Integer.parseInt(in.next());}
        long nl(){return Long.parseLong(in.next());}
        double nd(){return Double.parseDouble(in.next());}

        long mul(long a,long b){
            if(a>=mod)a%=mod;
            if(b>=mod)b%=mod;
            a*=b;
            if(a>=mod)a%=mod;
            return a;
        }
        long modPow(long a, long p){
            long o = 1;
            while(p>0){
                if((p&1)==1)o = mul(o,a);
                a = mul(a,a);
                p>>=1;
            }
            return o;
        }
        long add(long a, long b){
            if(a>=mod)a%=mod;
            if(b>=mod)b%=mod;
            if(b<0)b+=mod;
            a+=b;
            if(a>=mod)a-=mod;
            return a;
        }
    }

    class InputReader{
        BufferedReader br;
        StringTokenizer st;
        public InputReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }
 
        public InputReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }
 
        String next(){
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }
 
        String nextLine(){
            String str = "";
            try{
                str = br.readLine();
            }catch (IOException e){
                e.printStackTrace();
            }   
            return str;
        }
    }
}       