#include <bits/stdc++.h>
using namespace std;
const int MOD = 998244353;
const int N = 200005;
const double PI = 4 * atan(1);
const long long MAX = 9223372036854775807;
int dx[] = {1, -1, 0, 0, 1, -1, 1, -1};
int dy[] = {0, 0, 1, -1, 1, 1, -1, -1};
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
bool issquare(long long w) { return trunc(sqrt(w)) * trunc(sqrt(w)) == w; }
bool isprime(long long u) {
  for (long long i = 2; i <= (int)sqrt(u); i++) {
    if (u % i == 0) return 0;
  }
  return 1;
}
long long mod(long long to_mod, long long MOD) {
  to_mod %= MOD;
  while (to_mod < 0) to_mod += MOD;
  return to_mod % MOD;
}
long long moduloMultiplication(long long a, long long b, long long mod) {
  long long res = 0;
  a %= mod;
  while (b) {
    if (b & 1) res = (res + a) % mod;
    a = (2 * a) % mod;
    b >>= 1;
  }
  return res;
}
long long power(long long x, long long y) {
  long long res = 1;
  x = x;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long n, m, x, y, k, z, w, l, r, ans, res;
long long a[1005];
long long b[1005];
vector<long long> v;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  a[0] = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i + 1];
  }
  for (int i = 1; i <= n; i++) {
    b[i - 1] = a[i] - a[i - 1];
    bool cond = true;
    for (int j = 1; j <= n; j++) {
      if (a[j] != b[(j - 1) % i] + a[j - 1]) {
        cond = false;
      }
    }
    if (cond) v.push_back(i);
  }
  cout << v.size() << endl;
  for (auto t : v) {
    cout << t << " ";
  }
  return 0;
}

### Prompt

Please create a solution in CPP to the following problem:

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;
const int MOD = 998244353;
const int N = 200005;
const double PI = 4 * atan(1);
const long long MAX = 9223372036854775807;
int dx[] = {1, -1, 0, 0, 1, -1, 1, -1};
int dy[] = {0, 0, 1, -1, 1, 1, -1, -1};
long long gcd(long long a, long long b) { return (b == 0 ? a : gcd(b, a % b)); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
bool issquare(long long w) { return trunc(sqrt(w)) * trunc(sqrt(w)) == w; }
bool isprime(long long u) {
  for (long long i = 2; i <= (int)sqrt(u); i++) {
    if (u % i == 0) return 0;
  }
  return 1;
}
long long mod(long long to_mod, long long MOD) {
  to_mod %= MOD;
  while (to_mod < 0) to_mod += MOD;
  return to_mod % MOD;
}
long long moduloMultiplication(long long a, long long b, long long mod) {
  long long res = 0;
  a %= mod;
  while (b) {
    if (b & 1) res = (res + a) % mod;
    a = (2 * a) % mod;
    b >>= 1;
  }
  return res;
}
long long power(long long x, long long y) {
  long long res = 1;
  x = x;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
long long n, m, x, y, k, z, w, l, r, ans, res;
long long a[1005];
long long b[1005];
vector<long long> v;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  a[0] = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i + 1];
  }
  for (int i = 1; i <= n; i++) {
    b[i - 1] = a[i] - a[i - 1];
    bool cond = true;
    for (int j = 1; j <= n; j++) {
      if (a[j] != b[(j - 1) % i] + a[j - 1]) {
        cond = false;
      }
    }
    if (cond) v.push_back(i);
  }
  cout << v.size() << endl;
  for (auto t : v) {
    cout << t << " ";
  }
  return 0;
}

```