#include <bits/stdc++.h>
using namespace std;
bool a[(int)2e6 + 3];
int aa[1003];
void sieve() {
  a[2] = true;
  for (int i = 3; i < 2e6 + 3; i += 2) a[i] = true;
  for (int i = 3; i < 2004; i++)
    for (int j = 2; i * j < 2e6 + 3; j++) a[i * j] = false;
}
vector<int> ans;
int main() {
  sieve();
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> aa[i];
  sort(aa, aa + n);
  for (int i = 0; i < n; i++) {
    if (aa[i] == 1) ans.push_back(1);
  }
  if (ans.size()) {
    for (int i = ans.size(); i < n; i++)
      if (a[1 + aa[i]]) {
        ans.push_back(aa[i]);
        break;
      }
  }
  if (ans.size() > 1) {
    cout << ans.size() << endl;
    for (int i = 0; i < ans.size(); i++) cout << ans[i] << ' ';
    return 0;
  }
  if (ans.size()) ans.pop_back();
  for (int i = 0; i < n - 1; i++) {
    for (int j = i + 1; j < n; j++) {
      if (a[aa[i] + aa[j]]) {
        cout << 2 << endl;
        cout << aa[i] << ' ' << aa[j] << endl;
        return 0;
      }
    }
  }
  cout << 1 << endl << aa[0] << endl;
  return 0;
}

### 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 a[(int)2e6 + 3];
int aa[1003];
void sieve() {
  a[2] = true;
  for (int i = 3; i < 2e6 + 3; i += 2) a[i] = true;
  for (int i = 3; i < 2004; i++)
    for (int j = 2; i * j < 2e6 + 3; j++) a[i * j] = false;
}
vector<int> ans;
int main() {
  sieve();
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> aa[i];
  sort(aa, aa + n);
  for (int i = 0; i < n; i++) {
    if (aa[i] == 1) ans.push_back(1);
  }
  if (ans.size()) {
    for (int i = ans.size(); i < n; i++)
      if (a[1 + aa[i]]) {
        ans.push_back(aa[i]);
        break;
      }
  }
  if (ans.size() > 1) {
    cout << ans.size() << endl;
    for (int i = 0; i < ans.size(); i++) cout << ans[i] << ' ';
    return 0;
  }
  if (ans.size()) ans.pop_back();
  for (int i = 0; i < n - 1; i++) {
    for (int j = i + 1; j < n; j++) {
      if (a[aa[i] + aa[j]]) {
        cout << 2 << endl;
        cout << aa[i] << ' ' << aa[j] << endl;
        return 0;
      }
    }
  }
  cout << 1 << endl << aa[0] << endl;
  return 0;
}

```