#include <bits/stdc++.h>
using namespace std;
int main() {
  int n;
  cin >> n;
  vector<long long> a(n);
  long long max_ = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    max_ = max(max_, a[i]);
  }
  vector<bool> era(2 * max_ + 1);
  fill(era.begin(), era.end(), true);
  for (long long i = 2; i < era.size(); i++) {
    if (era[i]) {
      for (long long j = i * i; j < era.size(); j += i) {
        era[j] = false;
      }
    }
  }
  vector<vector<long long> > results;
  vector<long long> v(1);
  vector<long long> best;
  long long y = 0;
  bool flag;
  vector<long long> help;
  for (int i = 0; i < n; i++) {
    v[0] = a[i];
    results.push_back(v);
    if (1 > best.size()) {
      best = v;
    }
    y = results.size();
    for (long long j = 0; j < y - 1; j++) {
      flag = true;
      for (long long k = 0; k < results[j].size(); k++) {
        if (!era[results[j][k] + a[i]]) {
          flag = false;
          break;
        }
        if (results[j][k] == a[i]) {
          results[j].push_back(a[i]);
          if (results[j].size() > best.size()) {
            best = results[j];
          }
          flag = false;
          break;
        }
      }
      if (flag) {
        help = results[j];
        help.push_back(a[i]);
        results.push_back(help);
        if (help.size() > best.size()) {
          best = help;
        }
      }
    }
  }
  cout << best.size() << endl;
  for (long long i = 0; i < best.size(); i++) {
    cout << best[i] << " ";
  }
}

### Prompt

Please formulate 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;
int main() {
  int n;
  cin >> n;
  vector<long long> a(n);
  long long max_ = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    max_ = max(max_, a[i]);
  }
  vector<bool> era(2 * max_ + 1);
  fill(era.begin(), era.end(), true);
  for (long long i = 2; i < era.size(); i++) {
    if (era[i]) {
      for (long long j = i * i; j < era.size(); j += i) {
        era[j] = false;
      }
    }
  }
  vector<vector<long long> > results;
  vector<long long> v(1);
  vector<long long> best;
  long long y = 0;
  bool flag;
  vector<long long> help;
  for (int i = 0; i < n; i++) {
    v[0] = a[i];
    results.push_back(v);
    if (1 > best.size()) {
      best = v;
    }
    y = results.size();
    for (long long j = 0; j < y - 1; j++) {
      flag = true;
      for (long long k = 0; k < results[j].size(); k++) {
        if (!era[results[j][k] + a[i]]) {
          flag = false;
          break;
        }
        if (results[j][k] == a[i]) {
          results[j].push_back(a[i]);
          if (results[j].size() > best.size()) {
            best = results[j];
          }
          flag = false;
          break;
        }
      }
      if (flag) {
        help = results[j];
        help.push_back(a[i]);
        results.push_back(help);
        if (help.size() > best.size()) {
          best = help;
        }
      }
    }
  }
  cout << best.size() << endl;
  for (long long i = 0; i < best.size(); i++) {
    cout << best[i] << " ";
  }
}

```