#include <bits/stdc++.h>
int dx[] = {0, 0, -1, 1};
int dy[] = {1, -1, 0, 0};
using namespace std;
void file() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
bool isPrime(int x) {
  if (x % 2 == 0 && x != 2) return 0;
  if (x == 1) return 0;
  for (int i = 3; i * i <= x; i += 2)
    if (x % i == 0) return 0;
  return 1;
}
int main() {
  file();
  int n;
  cin >> n;
  vector<int> v(n);
  map<int, int> mp;
  for (auto& e : v) {
    cin >> e;
    mp[e]++;
  }
  if (mp[1] > 1) {
    vector<int> ans;
    while (mp[1]--) ans.push_back(1);
    for (int i = 0; i < n; i++) {
      if (isPrime(1 + v[i]) && v[i] != 1) {
        ans.push_back(v[i]);
        break;
      }
    }
    cout << int(ans.size()) << endl;
    for (auto e : ans) cout << e << " ";
    return 0;
  }
  for (int i = 0; i < n; i++) {
    for (int k = i + 1; k < n; k++) {
      if (isPrime(v[i] + v[k])) {
        return cout << 2 << endl << v[i] << " " << v[k], 0;
      }
    }
  }
  cout << 1 << endl << v[0] << endl;
}

### Prompt

Your challenge is to write 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>
int dx[] = {0, 0, -1, 1};
int dy[] = {1, -1, 0, 0};
using namespace std;
void file() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
bool isPrime(int x) {
  if (x % 2 == 0 && x != 2) return 0;
  if (x == 1) return 0;
  for (int i = 3; i * i <= x; i += 2)
    if (x % i == 0) return 0;
  return 1;
}
int main() {
  file();
  int n;
  cin >> n;
  vector<int> v(n);
  map<int, int> mp;
  for (auto& e : v) {
    cin >> e;
    mp[e]++;
  }
  if (mp[1] > 1) {
    vector<int> ans;
    while (mp[1]--) ans.push_back(1);
    for (int i = 0; i < n; i++) {
      if (isPrime(1 + v[i]) && v[i] != 1) {
        ans.push_back(v[i]);
        break;
      }
    }
    cout << int(ans.size()) << endl;
    for (auto e : ans) cout << e << " ";
    return 0;
  }
  for (int i = 0; i < n; i++) {
    for (int k = i + 1; k < n; k++) {
      if (isPrime(v[i] + v[k])) {
        return cout << 2 << endl << v[i] << " " << v[k], 0;
      }
    }
  }
  cout << 1 << endl << v[0] << endl;
}

```