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>
#pragma comment(linker, "/STACK:256000000")
std::uniform_int_distribution<std::mt19937::result_type> udist(0, 1);
std::mt19937 rng;
using namespace std;
FILE *in, *out;
int ri() {
  int a;
  fscanf(in, "%d", &a);
  return a;
}
double rf() {
  double a;
  fscanf(in, "%lf", &a);
  return a;
}
char sbuf[100005];
string rstr() {
  char c;
  char *b = sbuf;
  while (c = fgetc(in)) {
    if (c == '\n' || c == 255) break;
    *b++ = c;
  }
  *b = 0;
  return sbuf;
}
long long nod(long long a, long long b) {
  if (a == 1 || b == 1) return 1;
  if (a == 0) return b;
  if (b == 0) return a;
  return nod(b % a, a);
}
vector<int> primes;
void calculatePrimes() {
  primes.push_back(2);
  for (int i = 3; i < 1000000; i += 2) {
    bool bFound = false;
    int m = (int)sqrt((float)i);
    for (int j = 1; j < primes.size() && !bFound && primes[j] <= m; j++)
      bFound = ((i % primes[j]) == 0);
    if (!bFound) primes.push_back(i);
  }
}
bool isPrime(long long p) {
  for (int i : primes)
    if (p != i && (p % i) == 0) return false;
  return true;
}
void egcd(int a, int b, int &g, int &x, int &y) {
  if (a == 0) {
    g = b;
    x = 0;
    y = 1;
    return;
  }
  egcd(b % a, a, g, y, x);
  x = x - (b / a) * y;
}
int modinv(int a, int m) {
  int g = 0, x = 0, y = 0;
  egcd(a, m, g, x, y);
  return (x + m) % m;
}
long long modpow(int a, long long n, int p) {
  long long res = 1;
  long long curp = (a % p);
  while (n) {
    if (n & 1) {
      res = (res * curp) % p;
    }
    curp = (curp * curp) % p;
    n /= 2;
  }
  return res;
}
long long nf(int n) {
  if (n == 1) return 1;
  return (n * nf(n - 1)) % 998244353;
}
long long intpowmod(long long base, long long exp) {
  if (exp == 0) return 1;
  if (exp == 1) return base;
  if ((exp & 1) != 0)
    return (base * intpowmod((base * base) % 998244353, exp / 2)) % 998244353;
  return intpowmod((base * base) % 998244353, exp / 2);
}
int main() {
  long long n;
  cin >> n;
  map<int, set<int>> pm;
  int cvh, pr = 0;
  for (int i = (0); i < (((n))); ++i) {
    cin >> cvh;
    pm[cvh - pr].insert(i);
    pr = cvh;
  }
  vector<int> r;
  for (int i = 1; i <= n; i++) {
    bool can = true;
    for (auto &a : pm) {
      for (auto &b : a.second) {
        int next = b + i;
        if (next > n - 1) break;
        if (a.second.find(next) == a.second.end()) {
          can = false;
          break;
        }
      }
      if (!can) break;
    }
    if (can) r.push_back(i);
  }
  cout << r.size() << endl;
  for (auto &c : r) cout << c << " ";
  return 0;
}
