#include <bits/stdc++.h>
using namespace std;
const long long MAX_NUM = 2 * 1000000;
bool is_prime[MAX_NUM];
void init_primes() {
  fill(is_prime, is_prime + MAX_NUM, true);
  is_prime[0] = is_prime[1] = false;
  for (long long i = 2; i < MAX_NUM; i++) {
    if (is_prime[i]) {
      for (long long j = i * i; j < MAX_NUM; j += i) is_prime[j] = false;
    }
  }
}
void solve(vector<int>& v) {
  vector<int> odd, even;
  int cnt1 = 0;
  for (int i : v) {
    if (i & 1)
      odd.push_back(i);
    else
      even.push_back(i);
    if (i == 1) cnt1++;
  }
  if (cnt1 >= 1) {
    for (int e : even) {
      if (is_prime[e + 1]) {
        cout << cnt1 + 1 << "\n";
        for (int i = 0; i < cnt1; i++) cout << 1 << " ";
        cout << e << "\n";
        return;
      }
    }
  }
  if (cnt1 >= 2) {
    cout << cnt1 << "\n";
    for (int i = 0; i < cnt1; i++) cout << 1 << " ";
    cout << "\n";
    return;
  }
  for (int o : odd) {
    for (int e : even) {
      if (is_prime[o + e]) {
        cout << 2 << "\n";
        cout << o << " " << e << " "
             << "\n";
        return;
      }
    }
  }
  cout << 1 << "\n";
  cout << v[0] << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  init_primes();
  int N;
  cin >> N;
  vector<int> data(N, 0);
  for (int i = 0; i < N; i++) cin >> data[i];
  solve(data);
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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;
const long long MAX_NUM = 2 * 1000000;
bool is_prime[MAX_NUM];
void init_primes() {
  fill(is_prime, is_prime + MAX_NUM, true);
  is_prime[0] = is_prime[1] = false;
  for (long long i = 2; i < MAX_NUM; i++) {
    if (is_prime[i]) {
      for (long long j = i * i; j < MAX_NUM; j += i) is_prime[j] = false;
    }
  }
}
void solve(vector<int>& v) {
  vector<int> odd, even;
  int cnt1 = 0;
  for (int i : v) {
    if (i & 1)
      odd.push_back(i);
    else
      even.push_back(i);
    if (i == 1) cnt1++;
  }
  if (cnt1 >= 1) {
    for (int e : even) {
      if (is_prime[e + 1]) {
        cout << cnt1 + 1 << "\n";
        for (int i = 0; i < cnt1; i++) cout << 1 << " ";
        cout << e << "\n";
        return;
      }
    }
  }
  if (cnt1 >= 2) {
    cout << cnt1 << "\n";
    for (int i = 0; i < cnt1; i++) cout << 1 << " ";
    cout << "\n";
    return;
  }
  for (int o : odd) {
    for (int e : even) {
      if (is_prime[o + e]) {
        cout << 2 << "\n";
        cout << o << " " << e << " "
             << "\n";
        return;
      }
    }
  }
  cout << 1 << "\n";
  cout << v[0] << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  init_primes();
  int N;
  cin >> N;
  vector<int> data(N, 0);
  for (int i = 0; i < N; i++) cin >> data[i];
  solve(data);
  return 0;
}

```