#include <bits/stdc++.h>
using namespace std;
int dcmp(long double n, long double y) {
  return fabs(n - y) <= 1e-9 ? 0 : n < y ? -1 : 1;
}
const int N = 2e6 + 10;
bool prime[N];
int cnt[N];
vector<int> pr;
void seive() {
  for (int i = 2; i < N; i++) prime[i] = 1;
  prime[0] = prime[1] = 0;
  for (int i = 2; i * i < N; i++)
    if (prime[i])
      for (int j = i * i; j < N; j += i) prime[j] = 0;
  for (int i = 0; i < (int)(N); ++i)
    if (prime[i]) pr.push_back(i);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  seive();
  int n;
  cin >> n;
  vector<int> a(n);
  int cnt1 = 0;
  for (auto &x : a) {
    cin >> x;
    if (x == 1) cnt1++;
    cnt[x]++;
  }
  vector<int> ans;
  pair<int, int> idx = {-1, -1};
  int mx = -1;
  for (int i = 0; i < ((int)((a).size())); ++i) {
    for (int j = ((int)((pr).size())) - 1; j >= 0; j--) {
      int cur = 0;
      if (pr[j] - a[i] <= 0)
        continue;
      else {
        if (cnt[pr[j] - a[i]] > 0) {
          if (pr[j] - a[i] == a[i] && cnt[a[i]] == 1) continue;
          mx = 2;
          idx = {i, j};
        }
      }
    }
  }
  if (mx == 2) {
    ans.push_back(pr[idx.second] - a[idx.first]);
    ans.push_back(a[idx.first]);
  }
  for (int i = 0; i < ((int)((pr).size())); ++i) {
    if (cnt[pr[i] - 1] > 0 && pr[i] != 2 && cnt1 + 1 > mx) {
      ans.clear();
      mx = cnt1 + 1;
      for (int j = 0; j < (int)(cnt1); ++j) ans.push_back(1);
      ans.push_back(pr[i] - 1);
      break;
    }
  }
  if (cnt1 > mx) {
    ans.clear();
    for (int i = 0; i < (int)(cnt1); ++i) ans.push_back(1);
  }
  if (ans.size() == 0) ans.push_back(a[0]);
  cout << ans.size() << '\n';
  for (int i = 0; i < ((int)((ans).size())); ++i) {
    cout << ans[i] << ' ';
  }
  cout << '\n';
}

### Prompt

Create a solution in Cpp for 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 dcmp(long double n, long double y) {
  return fabs(n - y) <= 1e-9 ? 0 : n < y ? -1 : 1;
}
const int N = 2e6 + 10;
bool prime[N];
int cnt[N];
vector<int> pr;
void seive() {
  for (int i = 2; i < N; i++) prime[i] = 1;
  prime[0] = prime[1] = 0;
  for (int i = 2; i * i < N; i++)
    if (prime[i])
      for (int j = i * i; j < N; j += i) prime[j] = 0;
  for (int i = 0; i < (int)(N); ++i)
    if (prime[i]) pr.push_back(i);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  seive();
  int n;
  cin >> n;
  vector<int> a(n);
  int cnt1 = 0;
  for (auto &x : a) {
    cin >> x;
    if (x == 1) cnt1++;
    cnt[x]++;
  }
  vector<int> ans;
  pair<int, int> idx = {-1, -1};
  int mx = -1;
  for (int i = 0; i < ((int)((a).size())); ++i) {
    for (int j = ((int)((pr).size())) - 1; j >= 0; j--) {
      int cur = 0;
      if (pr[j] - a[i] <= 0)
        continue;
      else {
        if (cnt[pr[j] - a[i]] > 0) {
          if (pr[j] - a[i] == a[i] && cnt[a[i]] == 1) continue;
          mx = 2;
          idx = {i, j};
        }
      }
    }
  }
  if (mx == 2) {
    ans.push_back(pr[idx.second] - a[idx.first]);
    ans.push_back(a[idx.first]);
  }
  for (int i = 0; i < ((int)((pr).size())); ++i) {
    if (cnt[pr[i] - 1] > 0 && pr[i] != 2 && cnt1 + 1 > mx) {
      ans.clear();
      mx = cnt1 + 1;
      for (int j = 0; j < (int)(cnt1); ++j) ans.push_back(1);
      ans.push_back(pr[i] - 1);
      break;
    }
  }
  if (cnt1 > mx) {
    ans.clear();
    for (int i = 0; i < (int)(cnt1); ++i) ans.push_back(1);
  }
  if (ans.size() == 0) ans.push_back(a[0]);
  cout << ans.size() << '\n';
  for (int i = 0; i < ((int)((ans).size())); ++i) {
    cout << ans[i] << ' ';
  }
  cout << '\n';
}

```