#include <bits/stdc++.h>
using namespace std;
int const lim = 2e6 + 3;
int inp[lim + 3];
int arr[lim + 3];
vector<int> vec;
bool prime[lim + 3];
bool vis[lim + 3];
int done[lim + 3];
void solve() {
  for (int i = 2; i < 2 * lim; i++) {
    if (!vis[i]) {
      prime[i] = true;
      long long j = 1ll * i * i;
      while (j <= 2e6) vis[j] = true, j += i;
    }
  }
}
int main() {
  solve();
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> inp[i];
    done[inp[i]]++;
  }
  sort(inp, inp + n);
  vector<int> ans;
  if (done[1] >= 2) {
    int i = 0;
    while (i < n && inp[i] == 1) ans.push_back(inp[i]), i++;
    i = 0;
    for (int i = 0; i < n; i++) {
      if (inp[i] != 1 && prime[inp[i] + 1]) {
        ans.push_back(inp[i]);
        break;
      }
    }
  } else {
    bool shit = false;
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++)
        if (prime[inp[i] + inp[j]]) {
          ans.push_back(inp[i]), ans.push_back(inp[j]);
          shit = true;
          break;
        }
      if (shit) break;
    }
  }
  if (ans.size() == 0) {
    ans.push_back(inp[0]);
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
}

### 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;
int const lim = 2e6 + 3;
int inp[lim + 3];
int arr[lim + 3];
vector<int> vec;
bool prime[lim + 3];
bool vis[lim + 3];
int done[lim + 3];
void solve() {
  for (int i = 2; i < 2 * lim; i++) {
    if (!vis[i]) {
      prime[i] = true;
      long long j = 1ll * i * i;
      while (j <= 2e6) vis[j] = true, j += i;
    }
  }
}
int main() {
  solve();
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> inp[i];
    done[inp[i]]++;
  }
  sort(inp, inp + n);
  vector<int> ans;
  if (done[1] >= 2) {
    int i = 0;
    while (i < n && inp[i] == 1) ans.push_back(inp[i]), i++;
    i = 0;
    for (int i = 0; i < n; i++) {
      if (inp[i] != 1 && prime[inp[i] + 1]) {
        ans.push_back(inp[i]);
        break;
      }
    }
  } else {
    bool shit = false;
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++)
        if (prime[inp[i] + inp[j]]) {
          ans.push_back(inp[i]), ans.push_back(inp[j]);
          shit = true;
          break;
        }
      if (shit) break;
    }
  }
  if (ans.size() == 0) {
    ans.push_back(inp[0]);
  }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
}

```