#include <bits/stdc++.h>
using namespace std;
bool prime[2000005];
int n, ans, one;
vector<int> even, odd;
void seive() {
  for (int i = 2; i < 2000002; i++) {
    for (int j = 2 * i; j < 2000002; j = j + i) prime[j] = true;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int k;
    cin >> k;
    if (k == 1)
      one++;
    else if (k % 2 == 1)
      odd.push_back(k);
    else
      even.push_back(k);
  }
  seive();
  int st = -1;
  for (int i = 0; i < even.size(); i++) {
    if (prime[1 + even[i]] == false) {
      st = even[i];
      break;
    }
  }
  if (one > 1) {
    if (st != -1) {
      cout << one + 1 << endl;
      cout << st << " ";
    } else
      cout << one << endl;
    for (int i = 0; i < one; i++) cout << "1 ";
    cout << endl;
    return 0;
  } else if (one == 1) {
    if (st != -1) {
      cout << 2 << endl;
      cout << "1 " << st << endl;
      return 0;
    }
  }
  for (int i = 0; i < even.size(); i++) {
    for (int j = 0; j < odd.size(); j++) {
      if (prime[even[i] + odd[j]] == false) {
        cout << 2 << endl;
        cout << even[i] << " " << odd[j] << endl;
        return 0;
      }
    }
  }
  cout << 1 << endl;
  if (even.size() != 0)
    cout << even[0] << endl;
  else if (odd.size() != 0)
    cout << odd[0] << endl;
  else
    cout << 1 << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 prime[2000005];
int n, ans, one;
vector<int> even, odd;
void seive() {
  for (int i = 2; i < 2000002; i++) {
    for (int j = 2 * i; j < 2000002; j = j + i) prime[j] = true;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int k;
    cin >> k;
    if (k == 1)
      one++;
    else if (k % 2 == 1)
      odd.push_back(k);
    else
      even.push_back(k);
  }
  seive();
  int st = -1;
  for (int i = 0; i < even.size(); i++) {
    if (prime[1 + even[i]] == false) {
      st = even[i];
      break;
    }
  }
  if (one > 1) {
    if (st != -1) {
      cout << one + 1 << endl;
      cout << st << " ";
    } else
      cout << one << endl;
    for (int i = 0; i < one; i++) cout << "1 ";
    cout << endl;
    return 0;
  } else if (one == 1) {
    if (st != -1) {
      cout << 2 << endl;
      cout << "1 " << st << endl;
      return 0;
    }
  }
  for (int i = 0; i < even.size(); i++) {
    for (int j = 0; j < odd.size(); j++) {
      if (prime[even[i] + odd[j]] == false) {
        cout << 2 << endl;
        cout << even[i] << " " << odd[j] << endl;
        return 0;
      }
    }
  }
  cout << 1 << endl;
  if (even.size() != 0)
    cout << even[0] << endl;
  else if (odd.size() != 0)
    cout << odd[0] << endl;
  else
    cout << 1 << endl;
  return 0;
}

```