#include <bits/stdc++.h>
using namespace std;
int n;
vector<int> one, odd, even;
bool isPrime(int n) {
  if (n < 2) return false;
  if (n == 2) return true;
  if (n % 2 == 0) return false;
  int l = (int)(sqrt(n * 1.0));
  for (int i = 3; i <= l; i += 2)
    if (n % i == 0) return false;
  return true;
}
void tryPickOneEven(int& ans, vector<int>& solution) {
  if (one.size() == 0 && even.size() == 0) return;
  if (one.size() == 0) {
    ans = 1;
    solution.push_back(even[0]);
    return;
  }
  if (even.size() == 0) {
    ans = one.size();
    solution = one;
    return;
  }
  for (int i = 0; i < even.size(); ++i) {
    if (isPrime(even[i] + 1)) {
      ans = one.size() + 1;
      solution = one;
      solution.push_back(even[i]);
      return;
    }
  }
  ans = one.size();
  solution = one;
}
void tryPickOddEven(int& ans, vector<int>& solution) {
  if (odd.size() == 0 && even.size() == 0) return;
  if (odd.size() == 0) return;
  if (even.size() == 0) {
    if (ans == 0) {
      ans = 1;
      solution.push_back(odd[0]);
    }
    return;
  }
  if (ans >= 2) return;
  for (int i = 0; i < odd.size(); ++i)
    for (int j = 0; j < even.size(); ++j) {
      if (isPrime(odd[i] + even[j])) {
        ans = 2;
        solution.clear();
        solution.push_back(odd[i]);
        solution.push_back(even[j]);
        return;
      }
    }
}
int main() {
  cin >> n;
  int t;
  for (int i = 0; i < n; ++i) {
    cin >> t;
    if (t == 1)
      one.push_back(t);
    else if (t % 2 == 1)
      odd.push_back(t);
    else
      even.push_back(t);
  }
  int ans = 0;
  vector<int> solution;
  tryPickOneEven(ans, solution);
  tryPickOddEven(ans, solution);
  cout << ans << endl;
  cout << solution[0];
  for (int i = 1; i < solution.size(); ++i) cout << " " << solution[i];
  cout << endl;
  return 0;
}

### Prompt

Generate 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;
int n;
vector<int> one, odd, even;
bool isPrime(int n) {
  if (n < 2) return false;
  if (n == 2) return true;
  if (n % 2 == 0) return false;
  int l = (int)(sqrt(n * 1.0));
  for (int i = 3; i <= l; i += 2)
    if (n % i == 0) return false;
  return true;
}
void tryPickOneEven(int& ans, vector<int>& solution) {
  if (one.size() == 0 && even.size() == 0) return;
  if (one.size() == 0) {
    ans = 1;
    solution.push_back(even[0]);
    return;
  }
  if (even.size() == 0) {
    ans = one.size();
    solution = one;
    return;
  }
  for (int i = 0; i < even.size(); ++i) {
    if (isPrime(even[i] + 1)) {
      ans = one.size() + 1;
      solution = one;
      solution.push_back(even[i]);
      return;
    }
  }
  ans = one.size();
  solution = one;
}
void tryPickOddEven(int& ans, vector<int>& solution) {
  if (odd.size() == 0 && even.size() == 0) return;
  if (odd.size() == 0) return;
  if (even.size() == 0) {
    if (ans == 0) {
      ans = 1;
      solution.push_back(odd[0]);
    }
    return;
  }
  if (ans >= 2) return;
  for (int i = 0; i < odd.size(); ++i)
    for (int j = 0; j < even.size(); ++j) {
      if (isPrime(odd[i] + even[j])) {
        ans = 2;
        solution.clear();
        solution.push_back(odd[i]);
        solution.push_back(even[j]);
        return;
      }
    }
}
int main() {
  cin >> n;
  int t;
  for (int i = 0; i < n; ++i) {
    cin >> t;
    if (t == 1)
      one.push_back(t);
    else if (t % 2 == 1)
      odd.push_back(t);
    else
      even.push_back(t);
  }
  int ans = 0;
  vector<int> solution;
  tryPickOneEven(ans, solution);
  tryPickOddEven(ans, solution);
  cout << ans << endl;
  cout << solution[0];
  for (int i = 1; i < solution.size(); ++i) cout << " " << solution[i];
  cout << endl;
  return 0;
}

```