#include <bits/stdc++.h>
using namespace std;
void solve(int n) {
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  const int bound = *max_element((a).begin(), (a).end()) + 3;
  vector<int> cnt(bound);
  for (int x : a) cnt[x]++;
  vector<char> isprime(bound + bound, 1);
  isprime[0] = isprime[1] = 1;
  for (int i = 2; i < (int)(isprime).size(); i++)
    if (isprime[i])
      for (int j = 2 * i; j < (int)(isprime).size(); j += i) isprime[j] = 0;
  vector<int> res = {a[0]};
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++)
      if (isprime[a[i] + a[j]]) res = {a[i], a[j]};
  if (cnt[1] > 2) res = vector<int>(cnt[1], 1);
  for (int i = 0; i < n; i++)
    if (isprime[a[i] + 1] && a[i] != 1 && cnt[1] + 1 > (int)(res).size()) {
      res = vector<int>(cnt[1], 1);
      res.push_back(a[i]);
    }
  cout << (int)(res).size() << '\n';
  for (int x : res) cout << x << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  while (cin >> n) solve(n);
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
void solve(int n) {
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  const int bound = *max_element((a).begin(), (a).end()) + 3;
  vector<int> cnt(bound);
  for (int x : a) cnt[x]++;
  vector<char> isprime(bound + bound, 1);
  isprime[0] = isprime[1] = 1;
  for (int i = 2; i < (int)(isprime).size(); i++)
    if (isprime[i])
      for (int j = 2 * i; j < (int)(isprime).size(); j += i) isprime[j] = 0;
  vector<int> res = {a[0]};
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++)
      if (isprime[a[i] + a[j]]) res = {a[i], a[j]};
  if (cnt[1] > 2) res = vector<int>(cnt[1], 1);
  for (int i = 0; i < n; i++)
    if (isprime[a[i] + 1] && a[i] != 1 && cnt[1] + 1 > (int)(res).size()) {
      res = vector<int>(cnt[1], 1);
      res.push_back(a[i]);
    }
  cout << (int)(res).size() << '\n';
  for (int x : res) cout << x << ' ';
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  while (cin >> n) solve(n);
}

```