#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n;
  cin >> n;
  vector<bool> prime(3e6 + 12, true);
  for (long long i = 2; i < prime.size(); i++) {
    if (prime[i]) {
      for (long long j = i * 2; j < prime.size(); j += i) {
        prime[j] = false;
      }
    }
  }
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  vector<long long> ans;
  long long pos = -1;
  for (long long i = 0; i < n; i++) {
    if (a[i] == 1) {
      ans.push_back(1);
      pos = i;
    }
  }
  pos++;
  if (pos == n) {
    cout << ans.size() << "\n";
    for (auto u : ans) {
      cout << u << " ";
    }
    return;
  }
  bool good = false;
  for (long long i = pos; i < n; i++) {
    for (long long j = pos + 1; j < n; j++) {
      if (i == j) continue;
      if (prime[a[i] + a[j]] && !good) {
        bool ok = true;
        if (ans.size()) {
          if (!prime[a[i] + 1] || !prime[a[j] + 1]) {
            ok = false;
          }
        }
        if (ok) {
          ans.push_back(a[i]);
          ans.push_back(a[j]);
          good = true;
        }
      }
    }
  }
  if (!good) {
    for (long long i = pos; i < n; i++) {
      if (!ans.size() || prime[1 + a[i]]) {
        ans.push_back(a[i]);
        break;
      }
    }
  }
  vector<long long> ans2;
  for (long long i = 0; i < n; i++) {
    for (long long j = i + 1; j < n; j++) {
      if (prime[a[i] + a[j]]) {
        ans2.push_back(a[i]);
        ans2.push_back(a[j]);
        break;
      }
    }
    if (ans2.size()) break;
  }
  if (ans2.size() > ans.size()) swap(ans, ans2);
  cout << ans.size() << "\n";
  for (auto u : ans) {
    cout << u << ' ';
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout.precision(30);
  long long t = 1;
  while (t--) solve();
}

### Prompt

Generate 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;
void solve() {
  long long n;
  cin >> n;
  vector<bool> prime(3e6 + 12, true);
  for (long long i = 2; i < prime.size(); i++) {
    if (prime[i]) {
      for (long long j = i * 2; j < prime.size(); j += i) {
        prime[j] = false;
      }
    }
  }
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  vector<long long> ans;
  long long pos = -1;
  for (long long i = 0; i < n; i++) {
    if (a[i] == 1) {
      ans.push_back(1);
      pos = i;
    }
  }
  pos++;
  if (pos == n) {
    cout << ans.size() << "\n";
    for (auto u : ans) {
      cout << u << " ";
    }
    return;
  }
  bool good = false;
  for (long long i = pos; i < n; i++) {
    for (long long j = pos + 1; j < n; j++) {
      if (i == j) continue;
      if (prime[a[i] + a[j]] && !good) {
        bool ok = true;
        if (ans.size()) {
          if (!prime[a[i] + 1] || !prime[a[j] + 1]) {
            ok = false;
          }
        }
        if (ok) {
          ans.push_back(a[i]);
          ans.push_back(a[j]);
          good = true;
        }
      }
    }
  }
  if (!good) {
    for (long long i = pos; i < n; i++) {
      if (!ans.size() || prime[1 + a[i]]) {
        ans.push_back(a[i]);
        break;
      }
    }
  }
  vector<long long> ans2;
  for (long long i = 0; i < n; i++) {
    for (long long j = i + 1; j < n; j++) {
      if (prime[a[i] + a[j]]) {
        ans2.push_back(a[i]);
        ans2.push_back(a[j]);
        break;
      }
    }
    if (ans2.size()) break;
  }
  if (ans2.size() > ans.size()) swap(ans, ans2);
  cout << ans.size() << "\n";
  for (auto u : ans) {
    cout << u << ' ';
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout.precision(30);
  long long t = 1;
  while (t--) solve();
}

```