#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
int n, k;
vector<int> primes;
bool is_prime(int x) {
  for (int i = 0; i < primes.size() && primes[i] * primes[i] <= x; i++) {
    if (x % primes[i] == 0) return false;
  }
  return true;
}
void init() {
  for (int i = 2; i < 1500; i++) {
    if (is_prime(i)) {
      primes.push_back(i);
    }
  }
}
void solve() {
  init();
  cin >> n;
  vector<int> a(n);
  int s1 = 0;
  for (auto &i : a) {
    cin >> i;
    if (i == 1) s1++;
  }
  sort(a.begin(), a.end());
  vector<int> st;
  if (s1 > 0) {
    for (int i = 0; i < s1; i++) st.push_back(1);
    for (int i = 0; i < n; i++) {
      if (a[i] != 1 && is_prime(a[i] + 1)) {
        st.push_back(a[i]);
        break;
      }
    }
  }
  for (int i = 0; i < n && st.size() < 2; i++) {
    for (int j = i + 1; j < n; j++) {
      if (is_prime(a[i] + a[j])) {
        st.clear();
        st.push_back(a[i]);
        st.push_back(a[j]);
        break;
      }
    }
  }
  for (int i = 0; i < n && st.size() < 1; i++) {
    st.push_back(a[i]);
    break;
  }
  cout << st.size() << endl;
  for (int i = 0; i < st.size(); i++) {
    cout << st[i] << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
}

### Prompt

Please provide a cpp coded solution 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;
const long long mod = 1e9 + 7;
int n, k;
vector<int> primes;
bool is_prime(int x) {
  for (int i = 0; i < primes.size() && primes[i] * primes[i] <= x; i++) {
    if (x % primes[i] == 0) return false;
  }
  return true;
}
void init() {
  for (int i = 2; i < 1500; i++) {
    if (is_prime(i)) {
      primes.push_back(i);
    }
  }
}
void solve() {
  init();
  cin >> n;
  vector<int> a(n);
  int s1 = 0;
  for (auto &i : a) {
    cin >> i;
    if (i == 1) s1++;
  }
  sort(a.begin(), a.end());
  vector<int> st;
  if (s1 > 0) {
    for (int i = 0; i < s1; i++) st.push_back(1);
    for (int i = 0; i < n; i++) {
      if (a[i] != 1 && is_prime(a[i] + 1)) {
        st.push_back(a[i]);
        break;
      }
    }
  }
  for (int i = 0; i < n && st.size() < 2; i++) {
    for (int j = i + 1; j < n; j++) {
      if (is_prime(a[i] + a[j])) {
        st.clear();
        st.push_back(a[i]);
        st.push_back(a[j]);
        break;
      }
    }
  }
  for (int i = 0; i < n && st.size() < 1; i++) {
    st.push_back(a[i]);
    break;
  }
  cout << st.size() << endl;
  for (int i = 0; i < st.size(); i++) {
    cout << st[i] << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
}

```