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.
#include <bits/stdc++.h>
using namespace std;
const int INF = 2034567891;
const long long int INF64 = 4234567890123456789ll;
long long nc3(long long v) { return (v * (v - 1ll) * (v - 2ll)) / 6ll; }
long long nc2(long long v) { return (v * (v - 1ll)) / 2ll; }
long long int gcd(long long int a, long long int b) {
  return (b == 0 ? a : gcd(b, a % b));
}
long long int lcm(long long int a, long long int b) {
  return (a * (b / gcd(a, b)));
}
const int N = 10 + 2e5;
bool isprime(long long int n) {
  if (n < 2) return 0;
  if (n == 2) return 1;
  if (n % 2 == 0) return 0;
  for (long long int i = 3; i * i <= n; i++)
    if (n % i == 0) return 0;
  return 1;
}
double dist(pair<long long int, long long int> p1,
            pair<long long int, long long int> p2) {
  return (p1.first - p2.first) * (p1.first - p2.first) +
         (p1.second - p2.second) * (p1.second - p2.second);
}
long long int power(long long int x, long long int y) {
  long long int ans = 1;
  while (y > 0) {
    if (y & 1) ans = (ans * x);
    x = (x * x);
    y /= 2;
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int n;
  cin >> n;
  long long int a[n + 1];
  a[0] = 0;
  for (long long int i = 1; i <= n; i++) cin >> a[i];
  set<long long int> s;
  long long int len = 1;
  for (long long int i = 1; i <= n; i++) {
    long long int x[i];
    for (long long int j = 0; j < i; j++) x[j] = INF;
    long long int f = 0;
    for (long long int j = 1; j <= n; j++) {
      long long int tmp = a[j] - a[j - 1];
      long long int idx = (j - 1) % i;
      if (x[idx] != INF && x[idx] != tmp) {
        f = 1;
        break;
      }
      x[idx] = tmp;
    }
    if (!f) s.insert(i);
  }
  cout << s.size() << endl;
  for (auto i : s) cout << i << " ";
}
