#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 5;
int prime[N];
vector<int> a;
void seive() {
  prime[0] = prime[1] = 1;
  for (long long i = 2; i < N; i++) {
    if (prime[i]) continue;
    a.push_back(i);
    for (long long j = i * i; j < N; j += i) {
      prime[j] = 1;
    }
  }
}
vector<int> adj[N];
int main() {
  seive();
  int n, v;
  cin >> n;
  map<int, int> mp;
  for (int i = 0; i < n; i++) {
    cin >> v;
    mp[v]++;
  }
  vector<int> ans;
  if (mp.count(1)) {
    for (int i = 0; i < mp[1]; i++) {
      ans.push_back(1);
    }
    for (auto i : mp) {
      if (i.first != 1 && !prime[1 + i.first]) {
        ans.push_back(i.first);
        break;
      }
    }
    if (ans.size() >= 2) {
      cout << ans.size() << "\n";
      for (auto i : ans) {
        cout << i << " ";
      }
      return 0;
    }
  }
  for (auto i : mp) {
    for (auto j : mp) {
      if (!prime[i.first + j.first] && i.first != j.first) {
        cout << "2\n";
        cout << i.first << " " << j.first << "\n";
        return 0;
      }
    }
  }
  for (auto i : mp) {
    cout << "1\n";
    cout << i.first << "\n";
    return 0;
  }
  return 0;
}

### Prompt

Your task is to create 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 N = 2e6 + 5;
int prime[N];
vector<int> a;
void seive() {
  prime[0] = prime[1] = 1;
  for (long long i = 2; i < N; i++) {
    if (prime[i]) continue;
    a.push_back(i);
    for (long long j = i * i; j < N; j += i) {
      prime[j] = 1;
    }
  }
}
vector<int> adj[N];
int main() {
  seive();
  int n, v;
  cin >> n;
  map<int, int> mp;
  for (int i = 0; i < n; i++) {
    cin >> v;
    mp[v]++;
  }
  vector<int> ans;
  if (mp.count(1)) {
    for (int i = 0; i < mp[1]; i++) {
      ans.push_back(1);
    }
    for (auto i : mp) {
      if (i.first != 1 && !prime[1 + i.first]) {
        ans.push_back(i.first);
        break;
      }
    }
    if (ans.size() >= 2) {
      cout << ans.size() << "\n";
      for (auto i : ans) {
        cout << i << " ";
      }
      return 0;
    }
  }
  for (auto i : mp) {
    for (auto j : mp) {
      if (!prime[i.first + j.first] && i.first != j.first) {
        cout << "2\n";
        cout << i.first << " " << j.first << "\n";
        return 0;
      }
    }
  }
  for (auto i : mp) {
    cout << "1\n";
    cout << i.first << "\n";
    return 0;
  }
  return 0;
}

```