#include <bits/stdc++.h>
using namespace std;
bool arr[1000][1000], vis[1000001], check;
vector<int> sol, inds;
unsigned int mx;
int n, sz;
bool is_prime(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return 0;
  return 1;
}
void max_array(int ind) {
  inds.push_back(ind);
  for (int i = 0; i < n; i++)
    if (arr[ind][i]) {
      sz = inds.size();
      check = 1;
      for (int j = 0; j < sz; j++)
        if (!arr[inds[j]][i]) {
          check = 0;
          break;
        }
      if (check) max_array(i);
    }
  if (inds.size() > mx) {
    mx = inds.size();
    sol = inds;
  }
  inds.pop_back();
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int cnt_one = -1;
  cin >> n;
  int x[n];
  for (int i = 0; i < n; i++) {
    cin >> x[i];
    cnt_one += (x[i] == 1);
    if (!vis[x[i]])
      vis[x[i]] = 1;
    else {
      n--;
      i--;
    }
  }
  sort(x, x + n);
  for (int i = 0; i < n - 1; i++)
    for (int j = i + 1; j < n; j++)
      if (is_prime(x[i] + x[j])) arr[j][i] = arr[i][j] = 1;
  for (int i = 0; i < n; i++) max_array(i);
  if (x[sol[0]] != 1 && cnt_one + 1 > mx) {
    cout << cnt_one + 1 << '\n';
    for (int i = -1; i < cnt_one; i++) cout << 1 << ' ';
  } else {
    if (x[sol[0]] == 1) {
      cout << mx + cnt_one << '\n';
      for (int i = 0; i < cnt_one; i++) cout << 1 << ' ';
    } else
      cout << mx << '\n';
    for (unsigned int i = 0; i < sol.size(); i++) cout << x[sol[i]] << ' ';
  }
  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;
bool arr[1000][1000], vis[1000001], check;
vector<int> sol, inds;
unsigned int mx;
int n, sz;
bool is_prime(int x) {
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return 0;
  return 1;
}
void max_array(int ind) {
  inds.push_back(ind);
  for (int i = 0; i < n; i++)
    if (arr[ind][i]) {
      sz = inds.size();
      check = 1;
      for (int j = 0; j < sz; j++)
        if (!arr[inds[j]][i]) {
          check = 0;
          break;
        }
      if (check) max_array(i);
    }
  if (inds.size() > mx) {
    mx = inds.size();
    sol = inds;
  }
  inds.pop_back();
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int cnt_one = -1;
  cin >> n;
  int x[n];
  for (int i = 0; i < n; i++) {
    cin >> x[i];
    cnt_one += (x[i] == 1);
    if (!vis[x[i]])
      vis[x[i]] = 1;
    else {
      n--;
      i--;
    }
  }
  sort(x, x + n);
  for (int i = 0; i < n - 1; i++)
    for (int j = i + 1; j < n; j++)
      if (is_prime(x[i] + x[j])) arr[j][i] = arr[i][j] = 1;
  for (int i = 0; i < n; i++) max_array(i);
  if (x[sol[0]] != 1 && cnt_one + 1 > mx) {
    cout << cnt_one + 1 << '\n';
    for (int i = -1; i < cnt_one; i++) cout << 1 << ' ';
  } else {
    if (x[sol[0]] == 1) {
      cout << mx + cnt_one << '\n';
      for (int i = 0; i < cnt_one; i++) cout << 1 << ' ';
    } else
      cout << mx << '\n';
    for (unsigned int i = 0; i < sol.size(); i++) cout << x[sol[i]] << ' ';
  }
  return 0;
}

```