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 long long OO = (4e18) + 9;
const int MOD = (1e9) + 7;
const int oo = 2147483647;
const double EPS = 1e-8;
const double PI = acos(-1.0);
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
int dX[] = {1, 0, -1, 0, 1, -1, 1, -1};
int dY[] = {0, 1, 0, -1, 1, 1, -1, -1};
string abc = "abcdefghijklmnopqrstuvwxyz";
string vowels = "aeiou";
int dcmp(long double d1, long double d2) {
  return fabs(d1 - d2) <= EPS ? 0 : d1 > d2 ? 1 : -1;
}
long long gcd(long long x, long long y) { return !y ? x : gcd(y, x % y); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
int ord(int r, int c, int nCol) { return (nCol * r) + c + 1; }
pair<int, int> rord(int n, int nCol) {
  int r = n % nCol, c = n / nCol;
  if (n % nCol == 0) c--, r = nCol;
  return pair<int, int>(r, c);
}
long long poww(long long num, long long p, long long md = MOD) {
  num %= md;
  long long ret = 1;
  while (p > 0) {
    if (p & 1) ret = (ret * num) % md;
    num = (num * num) % md;
    p >>= 1;
  }
  return ret;
}
long long nCr(long long n, long long r) {
  if (n < r) return 0;
  if (r == 0) return 1;
  return n * nCr(n - 1, r - 1) / r;
}
int a[10000];
int main() {
  int n, first = 1;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  vector<int> ans;
  ans.push_back(n);
  for (int i = 1; i < n; i++) {
    first = 1;
    for (int j = i; j < n; j++) {
      if (j % i == 0) {
        if (a[j] - a[j - 1] != a[0]) first = 0;
      } else {
        if (a[j] - a[j - 1] != a[j % i] - a[(j - 1) % i]) first = 0;
      }
    }
    if (first) ans.push_back(i);
  }
  cout << (int)ans.size() << endl;
  sort(ans.begin(), ans.end());
  for (int i = 0; i < (int)ans.size(); i++) cout << ans[i] << ' ';
  return 0;
}
