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.*;

public class Main {
	
	public static int mod = 1000000007;
	public static boolean check(int a[],int n, int d) {
		for(int i = 0; i+d<n; i++) {
			if(a[i+1] - a[i]  != a[i+d+1] - a[i+d]) return false;
		}
		return true;
	}
	public static void solve(InputReader in) {
		int n = in.readInt();
		int a[] = new int[n+1]; 
		
		for(int i = 1; i<=n; i++)
			a[i] = in.readInt();
		
		ArrayList<Integer> ans = new ArrayList<>();
		for(int i = 1; i<=n; i++) {
			if(check(a , n , i)) ans.add(i); 
		}
		System.out.println(ans.size()); 
		for(int i : ans) System.out.print(i + " "); System.out.println();
	}

	public static void main(String[] args) {
		InputReader in = new InputReader(System.in);
		int t =1;
		while (t-- > 0)
			solve(in);
	}
}


class InputReader{
	private InputStream stream;
	private byte[] buf = new byte[1024];
	private int curChar;
	private int numChars;
	private SpaceCharFilter filter;

	 public InputReader(InputStream stream) {
		this.stream = stream;
	}

     public int read() {
		if (numChars == -1)
			throw new InputMismatchException();
		if (curChar >= numChars) {
			curChar = 0;
			try {
				numChars = stream.read(buf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (numChars <= 0)
				return -1;
		}
		return buf[curChar++];
	}

     public int readInt() {
		int c = read();
		while (isSpaceChar(c))
			c = read();
		int sgn = 1;
		if (c == '-') {
			sgn = -1;
			c = read();
		}
		int res = 0;
		do {
			if (c < '0' || c > '9')
				throw new InputMismatchException();
			res *= 10;
			res += c - '0';
			c = read();
		} while (!isSpaceChar(c));
		return res * sgn;
	}

     public String readString() {
		int c = read();
		while (isSpaceChar(c))
			c = read();
		StringBuilder res = new StringBuilder();
		do {
			res.appendCodePoint(c);
			c = read();
		} while (!isSpaceChar(c));
		return res.toString();
	}

     public long readLong() {
		int c = read();
		while (isSpaceChar(c)) {
			c = read();
		}
		int sgn = 1;
		if (c == '-') {
			sgn = -1;
			c = read();
		}
		long res = 0;
		do {
			if (c < '0' || c > '9') {
				throw new InputMismatchException();
			}
			res *= 10;
			res += c - '0';
			c = read();
		} while (!isSpaceChar(c));
		return res * sgn;
	}
	
     public boolean isSpaceChar(int c) {
		if (filter != null)
			return filter.isSpaceChar(c);
		return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
	}
     
     public String next() {
		return readString();
	}
    
     public interface SpaceCharFilter {
		public boolean isSpaceChar(int ch);
	}
}
