#include <bits/stdc++.h>
using namespace std;
bool isprime(int n);
int main() {
  std::ios::sync_with_stdio(false);
  int n, t;
  set<int> odd, even;
  while (cin >> n) {
    int one = 0, flag = 0;
    for (int i = 0; i < n; i++) {
      cin >> t;
      if (t == 1) one++;
      if (t % 2 == 0)
        even.insert(t);
      else
        odd.insert(t);
    }
    if (one < 2) {
      for (set<int>::iterator it = odd.begin(); it != odd.end(); it++) {
        for (set<int>::iterator it2 = even.begin(); it2 != even.end(); it2++) {
          if (isprime(*it + *it2)) {
            cout << 2 << endl;
            cout << *it << " " << *it2 << endl;
            flag = 1;
            break;
          }
        }
        if (flag) break;
      }
      if (!flag) {
        cout << 1 << endl;
        if (one)
          cout << 1 << endl;
        else if (!odd.empty())
          cout << *odd.begin() << endl;
        else
          cout << *even.begin() << endl;
      }
    } else {
      for (set<int>::iterator it = even.begin(); it != even.end(); it++) {
        if (isprime(*it + 1)) {
          cout << one + 1 << endl;
          cout << *it;
          flag = 1;
          break;
        }
      }
      if (!flag) cout << one << endl;
      for (int i = 0; i < one; i++) {
        if (flag || i) cout << " ";
        cout << 1;
      }
      cout << endl;
      flag = 1;
    }
    odd.clear(), even.clear();
  }
  return 0;
}
bool isprime(int n) {
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return false;
  }
  return true;
}

### Prompt

Create a solution in CPP for 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;
bool isprime(int n);
int main() {
  std::ios::sync_with_stdio(false);
  int n, t;
  set<int> odd, even;
  while (cin >> n) {
    int one = 0, flag = 0;
    for (int i = 0; i < n; i++) {
      cin >> t;
      if (t == 1) one++;
      if (t % 2 == 0)
        even.insert(t);
      else
        odd.insert(t);
    }
    if (one < 2) {
      for (set<int>::iterator it = odd.begin(); it != odd.end(); it++) {
        for (set<int>::iterator it2 = even.begin(); it2 != even.end(); it2++) {
          if (isprime(*it + *it2)) {
            cout << 2 << endl;
            cout << *it << " " << *it2 << endl;
            flag = 1;
            break;
          }
        }
        if (flag) break;
      }
      if (!flag) {
        cout << 1 << endl;
        if (one)
          cout << 1 << endl;
        else if (!odd.empty())
          cout << *odd.begin() << endl;
        else
          cout << *even.begin() << endl;
      }
    } else {
      for (set<int>::iterator it = even.begin(); it != even.end(); it++) {
        if (isprime(*it + 1)) {
          cout << one + 1 << endl;
          cout << *it;
          flag = 1;
          break;
        }
      }
      if (!flag) cout << one << endl;
      for (int i = 0; i < one; i++) {
        if (flag || i) cout << " ";
        cout << 1;
      }
      cout << endl;
      flag = 1;
    }
    odd.clear(), even.clear();
  }
  return 0;
}
bool isprime(int n) {
  for (int i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return false;
  }
  return true;
}

```