#include <bits/stdc++.h>
using namespace std;
const int NMAX = 2000005;
vector<int> a;
vector<int> v;
vector<int> sol;
int c[NMAX];
void ciur(int n) {
  c[0] = c[1] = 1;
  for (int i = 4; i <= n; i += 2) c[i] = 1;
  int lim = (int)sqrt((double)n);
  for (int i = 3; i <= lim; i += 2)
    if (c[i] == 0)
      for (int j = i * i; j <= n; j += 2 * i) c[j] = 1;
}
bool verif(int x) {
  for (int i = 0; i < v.size(); i++)
    if (c[v[i] + x]) return false;
  return true;
}
int main() {
  ciur(2000000);
  int n, x;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    a.push_back(x);
  }
  sort(a.begin(), a.end());
  int ma = 0;
  for (int i = 0; i < a.size(); i++) {
    v.clear();
    for (int j = i; j < a.size(); j++) {
      if (v.empty()) {
        v.push_back(a[j]);
        continue;
      }
      if (verif(a[j])) v.push_back(a[j]);
    }
    if (v.size() > ma) {
      ma = v.size();
      sol.clear();
      for (int j = 0; j < v.size(); j++) sol.push_back(v[j]);
      if (ma > a.size() - i) break;
    }
  }
  cout << sol.size() << "\n";
  for (int i = 0; i < sol.size(); i++) cout << sol[i] << " ";
  cout << "\n";
  return 0;
}

### Prompt

Please formulate 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;
const int NMAX = 2000005;
vector<int> a;
vector<int> v;
vector<int> sol;
int c[NMAX];
void ciur(int n) {
  c[0] = c[1] = 1;
  for (int i = 4; i <= n; i += 2) c[i] = 1;
  int lim = (int)sqrt((double)n);
  for (int i = 3; i <= lim; i += 2)
    if (c[i] == 0)
      for (int j = i * i; j <= n; j += 2 * i) c[j] = 1;
}
bool verif(int x) {
  for (int i = 0; i < v.size(); i++)
    if (c[v[i] + x]) return false;
  return true;
}
int main() {
  ciur(2000000);
  int n, x;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    a.push_back(x);
  }
  sort(a.begin(), a.end());
  int ma = 0;
  for (int i = 0; i < a.size(); i++) {
    v.clear();
    for (int j = i; j < a.size(); j++) {
      if (v.empty()) {
        v.push_back(a[j]);
        continue;
      }
      if (verif(a[j])) v.push_back(a[j]);
    }
    if (v.size() > ma) {
      ma = v.size();
      sol.clear();
      for (int j = 0; j < v.size(); j++) sol.push_back(v[j]);
      if (ma > a.size() - i) break;
    }
  }
  cout << sol.size() << "\n";
  for (int i = 0; i < sol.size(); i++) cout << sol[i] << " ";
  cout << "\n";
  return 0;
}

```