#include <bits/stdc++.h>
using namespace std;
int arr[500009];
int rr[500009 * 10];
vector<int> v;
int main() {
  int n;
  int ans = 0;
  cin >> n;
  for (int i = 2; i <= 2000000; i++) {
    for (int j = i * 2; j <= 2000000; j += i) {
      rr[j] = 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  for (int i = 1; i <= n; i++) {
    if (arr[i] == 1) {
      ans++;
    }
  }
  if (ans < 2) {
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        if (arr[i] % 2 == 0 && arr[j] % 2 == 1) {
          if (!rr[arr[i] + arr[j]]) {
            cout << 2 << endl;
            cout << arr[i] << " " << arr[j] << endl;
            return 0;
          }
        }
      }
    }
  } else {
    for (int i = 1; i <= n; i++) {
      if (arr[i] % 2 == 0 && rr[arr[i] + 1] == 0) {
        cout << ans + 1 << endl;
        for (int i = 1; i <= ans; i++) cout << 1 << " ";
        cout << arr[i] << endl;
        return 0;
      }
    }
  }
  if (ans) {
    cout << ans << endl;
    for (int i = 1; i <= ans; i++) {
      cout << 1 << " ";
    }
  } else
    cout << 1 << endl << arr[1] << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp 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 arr[500009];
int rr[500009 * 10];
vector<int> v;
int main() {
  int n;
  int ans = 0;
  cin >> n;
  for (int i = 2; i <= 2000000; i++) {
    for (int j = i * 2; j <= 2000000; j += i) {
      rr[j] = 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  for (int i = 1; i <= n; i++) {
    if (arr[i] == 1) {
      ans++;
    }
  }
  if (ans < 2) {
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        if (arr[i] % 2 == 0 && arr[j] % 2 == 1) {
          if (!rr[arr[i] + arr[j]]) {
            cout << 2 << endl;
            cout << arr[i] << " " << arr[j] << endl;
            return 0;
          }
        }
      }
    }
  } else {
    for (int i = 1; i <= n; i++) {
      if (arr[i] % 2 == 0 && rr[arr[i] + 1] == 0) {
        cout << ans + 1 << endl;
        for (int i = 1; i <= ans; i++) cout << 1 << " ";
        cout << arr[i] << endl;
        return 0;
      }
    }
  }
  if (ans) {
    cout << ans << endl;
    for (int i = 1; i <= ans; i++) {
      cout << 1 << " ";
    }
  } else
    cout << 1 << endl << arr[1] << endl;
  return 0;
}

```