#include <bits/stdc++.h>
using namespace std;
int dx[8] = {0, 0, 1, -1, 1, 1, -1, -1};
int dy[8] = {1, -1, 0, 0, 1, -1, 1, -1};
template <typename T>
inline T BigMod(T A, T B) {
  T ret = 1;
  while (B) {
    if (B & 1) ret = (ret * A) % 1000000007;
    A = (A * A) % 1000000007;
    B = B >> 1;
  }
  return ret;
}
template <typename T>
inline T InvMod(T A, T M = 1000000007) {
  return BigMod(A, M - 2);
}
string tostring(long long int res) {
  if (!res) return "0";
  string curstr = "";
  while (res != 0) {
    long long int temp = (res % 10);
    curstr += ((char)temp + '0');
    res /= 10;
  }
  reverse(curstr.begin(), curstr.end());
  return curstr;
}
long long int toint(string ss) {
  long long int inss = 0;
  for (int i = 0; i < ss.size(); i++) {
    inss = (inss * 10) + ((int)(ss[i] - '0'));
  }
  return inss;
}
int n, k, ar[1003];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &ar[i]);
  ar[0] = 0;
  set<int> sst;
  for (int i = 1; i <= n; i++) {
    int temp = ar[i] - ar[i - 1];
    sst.insert(temp);
  }
  int fl = 0;
  vector<int> res;
  for (k = 1; k <= n; k++) {
    fl = 1;
    vector<int> st;
    for (int i = 1; i <= n; i++) {
      int temp = (ar[i] - ar[i - 1]);
      if (st.size() < k) {
        if (st.size() < k) st.push_back(temp);
      } else {
        int now = i % k;
        if (now == 0)
          now = st.size() - 1;
        else
          now--;
        temp = st[now] + ar[i - 1];
        if (temp != ar[i]) {
          fl = 0;
        }
      }
    }
    if (fl) res.push_back(k);
  }
  printf("%d\n", res.size());
  for (int i = 0; i < res.size(); i++) {
    if (!i)
      printf("%d", res[i]);
    else
      printf(" %d", res[i]);
  }
  printf("\n");
  return 0;
}

### Prompt

Please formulate a Cpp solution 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;
int dx[8] = {0, 0, 1, -1, 1, 1, -1, -1};
int dy[8] = {1, -1, 0, 0, 1, -1, 1, -1};
template <typename T>
inline T BigMod(T A, T B) {
  T ret = 1;
  while (B) {
    if (B & 1) ret = (ret * A) % 1000000007;
    A = (A * A) % 1000000007;
    B = B >> 1;
  }
  return ret;
}
template <typename T>
inline T InvMod(T A, T M = 1000000007) {
  return BigMod(A, M - 2);
}
string tostring(long long int res) {
  if (!res) return "0";
  string curstr = "";
  while (res != 0) {
    long long int temp = (res % 10);
    curstr += ((char)temp + '0');
    res /= 10;
  }
  reverse(curstr.begin(), curstr.end());
  return curstr;
}
long long int toint(string ss) {
  long long int inss = 0;
  for (int i = 0; i < ss.size(); i++) {
    inss = (inss * 10) + ((int)(ss[i] - '0'));
  }
  return inss;
}
int n, k, ar[1003];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &ar[i]);
  ar[0] = 0;
  set<int> sst;
  for (int i = 1; i <= n; i++) {
    int temp = ar[i] - ar[i - 1];
    sst.insert(temp);
  }
  int fl = 0;
  vector<int> res;
  for (k = 1; k <= n; k++) {
    fl = 1;
    vector<int> st;
    for (int i = 1; i <= n; i++) {
      int temp = (ar[i] - ar[i - 1]);
      if (st.size() < k) {
        if (st.size() < k) st.push_back(temp);
      } else {
        int now = i % k;
        if (now == 0)
          now = st.size() - 1;
        else
          now--;
        temp = st[now] + ar[i - 1];
        if (temp != ar[i]) {
          fl = 0;
        }
      }
    }
    if (fl) res.push_back(k);
  }
  printf("%d\n", res.size());
  for (int i = 0; i < res.size(); i++) {
    if (!i)
      printf("%d", res[i]);
    else
      printf(" %d", res[i]);
  }
  printf("\n");
  return 0;
}

```