#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T SUM(const vector<T>& v) {
  return accumulate(begin(v), end(v), T(0), [](T a, T b) { return a + b; });
}
bool isPrime(unsigned long long n) {
  if (n == 0 || n == 1) return false;
  if (n == 2 || n == 3 || n == 5 || n == 7) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i += 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  vector<int> arr(N);
  int nb_ones = 0;
  int cand = -1;
  for (int i = 0; i < N; i++) {
    cin >> arr[i];
    if (arr[i] == 1) nb_ones++;
    if (isPrime(arr[i] + 1) && arr[i] != 1) cand = arr[i];
  }
  int tot = nb_ones;
  if (nb_ones > 0 && cand != -1) tot++;
  if (tot >= 2) {
    cout << tot << endl;
    arr = vector<int>(nb_ones, 1);
    if (cand != -1) arr.push_back(cand);
    {
      copy(begin(arr), end(arr) - 1, ostream_iterator<int>(cout, " "));
      cout << *(end(arr) - 1) << endl;
    };
    return 0;
  }
  for (int i = 0; i < N; i++) {
    for (int j = i + 1; j < N; j++) {
      if (isPrime(arr[i] + arr[j])) {
        cout << 2 << endl << arr[i] << " " << arr[j] << endl;
        return 0;
      }
    }
  }
  cout << 1 << endl << arr[0] << endl;
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

A tuple of positive integers {x1, x2, ..., xk} is called simple if for all pairs of positive integers (i, j) (1 ≤ i < j ≤ k), xi + xj is a prime.

You are given an array a with n positive integers a1, a2, ..., an (not necessary distinct). You want to find a simple subset of the array a with the maximum size.

A prime number (or a prime) is a natural number greater than 1 that has no positive divisors other than 1 and itself.

Let's define a subset of the array a as a tuple that can be obtained from a by removing some (possibly all) elements of it.

Input

The first line contains integer n (1 ≤ n ≤ 1000) — the number of integers in the array a.

The second line contains n integers ai (1 ≤ ai ≤ 106) — the elements of the array a.

Output

On the first line print integer m — the maximum possible size of simple subset of a.

On the second line print m integers bl — the elements of the simple subset of the array a with the maximum size.

If there is more than one solution you can print any of them. You can print the elements of the subset in any order.

Examples

Input

2
2 3


Output

2
3 2


Input

2
2 2


Output

1
2


Input

3
2 1 1


Output

3
1 1 2


Input

2
83 14


Output

2
14 83

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline T SUM(const vector<T>& v) {
  return accumulate(begin(v), end(v), T(0), [](T a, T b) { return a + b; });
}
bool isPrime(unsigned long long n) {
  if (n == 0 || n == 1) return false;
  if (n == 2 || n == 3 || n == 5 || n == 7) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i += 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  vector<int> arr(N);
  int nb_ones = 0;
  int cand = -1;
  for (int i = 0; i < N; i++) {
    cin >> arr[i];
    if (arr[i] == 1) nb_ones++;
    if (isPrime(arr[i] + 1) && arr[i] != 1) cand = arr[i];
  }
  int tot = nb_ones;
  if (nb_ones > 0 && cand != -1) tot++;
  if (tot >= 2) {
    cout << tot << endl;
    arr = vector<int>(nb_ones, 1);
    if (cand != -1) arr.push_back(cand);
    {
      copy(begin(arr), end(arr) - 1, ostream_iterator<int>(cout, " "));
      cout << *(end(arr) - 1) << endl;
    };
    return 0;
  }
  for (int i = 0; i < N; i++) {
    for (int j = i + 1; j < N; j++) {
      if (isPrime(arr[i] + arr[j])) {
        cout << 2 << endl << arr[i] << " " << arr[j] << endl;
        return 0;
      }
    }
  }
  cout << 1 << endl << arr[0] << endl;
  return 0;
}

```