#include <bits/stdc++.h>
using namespace std;
int sim(int k) {
  int n = 2;
  while (n * n <= k) {
    if (k % n == 0) return 0;
    n++;
  }
  return 1;
}
int main() {
  int n;
  cin >> n;
  int m1 = 0, me = 0, mo = 0, a;
  vector<int> e;
  vector<int> odd;
  for (int i = 0; i < n; i++) {
    cin >> a;
    if (a % 2 == 0) {
      e.push_back(a);
    } else {
      if (a == 1) {
        m1++;
      } else {
        odd.push_back(a);
      }
    }
  }
  me = e.size();
  mo = odd.size();
  if (m1 == 0) {
    for (int i = 0; i < mo; i++)
      for (int j = 0; j < me; j++) {
        if (sim(e[j] + odd[i])) {
          cout << 2 << endl;
          cout << e[j] << " " << odd[i];
          return 0;
        }
      }
    cout << 1 << endl;
    cout << a;
  } else {
    for (int i = 0; i < me; i++) {
      if (sim(e[i] + 1)) {
        cout << m1 + 1 << endl;
        cout << e[i] << " ";
        for (int j = 0; j < m1; j++) {
          cout << 1 << " ";
        }
        return 0;
      }
    }
    if (m1 == 1) {
      for (int i = 0; i < mo; i++)
        for (int j = 0; j < me; j++) {
          if (sim(e[j] + odd[i])) {
            cout << 2 << endl;
            cout << e[j] << " " << odd[i];
            return 0;
          }
        }
    }
    cout << m1 << endl;
    for (int j = 0; j < m1; j++) {
      cout << 1 << " ";
    }
    return 0;
  }
  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;
int sim(int k) {
  int n = 2;
  while (n * n <= k) {
    if (k % n == 0) return 0;
    n++;
  }
  return 1;
}
int main() {
  int n;
  cin >> n;
  int m1 = 0, me = 0, mo = 0, a;
  vector<int> e;
  vector<int> odd;
  for (int i = 0; i < n; i++) {
    cin >> a;
    if (a % 2 == 0) {
      e.push_back(a);
    } else {
      if (a == 1) {
        m1++;
      } else {
        odd.push_back(a);
      }
    }
  }
  me = e.size();
  mo = odd.size();
  if (m1 == 0) {
    for (int i = 0; i < mo; i++)
      for (int j = 0; j < me; j++) {
        if (sim(e[j] + odd[i])) {
          cout << 2 << endl;
          cout << e[j] << " " << odd[i];
          return 0;
        }
      }
    cout << 1 << endl;
    cout << a;
  } else {
    for (int i = 0; i < me; i++) {
      if (sim(e[i] + 1)) {
        cout << m1 + 1 << endl;
        cout << e[i] << " ";
        for (int j = 0; j < m1; j++) {
          cout << 1 << " ";
        }
        return 0;
      }
    }
    if (m1 == 1) {
      for (int i = 0; i < mo; i++)
        for (int j = 0; j < me; j++) {
          if (sim(e[j] + odd[i])) {
            cout << 2 << endl;
            cout << e[j] << " " << odd[i];
            return 0;
          }
        }
    }
    cout << m1 << endl;
    for (int j = 0; j < m1; j++) {
      cout << 1 << " ";
    }
    return 0;
  }
  return 0;
}

```