#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int modInverse(long long int n, long long int p) {
  return power(n, p - 2, p);
}
struct point {
  long long int x, y;
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  vector<long long int> arr(n + 1);
  for (long long int i = 1; i < n + 1; i++) {
    cin >> arr[i];
  }
  vector<long long int> ans;
  for (long long int k = 1; k < n + 1; k++) {
    long long int sum = 0;
    bool flag = true;
    vector<long long int> x(k);
    unordered_map<long long int, long long int> ps;
    for (long long int i = 1; i < n + 1; i++) {
      if (i <= k) {
        x[i - 1] = arr[i] - ps[i - 2];
        ps[i - 1] = ps[i - 2] + x[i - 1];
        sum += x[i - 1];
      } else {
        long long int ind = i / k;
        long long int ind1 = i % k - 1;
        long long int cur = ind * sum + ps[ind1];
        if (cur != arr[i]) {
          flag = false;
        }
      }
    }
    if (flag) {
      ans.push_back(k);
    }
  }
  cout << ans.size() << "\n";
  for (auto x : ans) {
    cout << x << " ";
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long int modInverse(long long int n, long long int p) {
  return power(n, p - 2, p);
}
struct point {
  long long int x, y;
};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  vector<long long int> arr(n + 1);
  for (long long int i = 1; i < n + 1; i++) {
    cin >> arr[i];
  }
  vector<long long int> ans;
  for (long long int k = 1; k < n + 1; k++) {
    long long int sum = 0;
    bool flag = true;
    vector<long long int> x(k);
    unordered_map<long long int, long long int> ps;
    for (long long int i = 1; i < n + 1; i++) {
      if (i <= k) {
        x[i - 1] = arr[i] - ps[i - 2];
        ps[i - 1] = ps[i - 2] + x[i - 1];
        sum += x[i - 1];
      } else {
        long long int ind = i / k;
        long long int ind1 = i % k - 1;
        long long int cur = ind * sum + ps[ind1];
        if (cur != arr[i]) {
          flag = false;
        }
      }
    }
    if (flag) {
      ans.push_back(k);
    }
  }
  cout << ans.size() << "\n";
  for (auto x : ans) {
    cout << x << " ";
  }
}

```