#include <bits/stdc++.h>
using namespace std;
const int N = (int)2e6 + 10;
int n;
bool is_prime[N];
vector<int> a;
vector<pair<int, int> > g[N];
set<int> was;
void build_prime() {
  for (int i = 2; i < N; ++i) is_prime[i] = true;
  for (int i = 2; i < N; ++i) {
    if (is_prime[i]) {
      int j = i + i;
      while (j < N) {
        is_prime[j] = false;
        j += i;
      }
    }
  }
}
int cnt1;
vector<int> ans1, ans;
int main() {
  build_prime();
  cin >> n;
  srand(time(NULL));
  for (int i = 0; i < n; ++i) {
    int x;
    cin >> x;
    if (x == 1) ans1.push_back(1);
    if (x == 1 || !was.count(x)) a.push_back(x);
    was.insert(x);
  }
  was.clear();
  for (int i = 0; i < a.size(); ++i)
    for (int j = 0; j < a.size(); ++j) {
      if (i == j) continue;
      if (is_prime[a[i] + a[j]]) {
        if (ans.size() == 0) {
          ans.push_back(a[i]);
          ans.push_back(a[j]);
        }
        if (a[i] != 1 && a[j] != 1 && is_prime[a[i] + 1] &&
            is_prime[a[j] + 1] && (ans1.size() != 0 || ans1.back() == 1)) {
          ans1.push_back(a[i]);
          ans1.push_back(a[j]);
        }
      }
    }
  if (ans.size() == 0) {
    for (int i = 0; i < a.size(); ++i) {
      ans.push_back(a[i]);
      break;
    }
  }
  if (ans1.size() != 0 && ans1.back() == 1) {
    for (int i = 0; i < a.size(); ++i) {
      if (is_prime[a[i] + 1] && a[i] != 1) {
        ans1.push_back(a[i]);
        break;
      }
    }
  }
  if (ans1.size() > ans.size()) {
    cout << ans1.size() << endl;
    for (int i = 0; i < ans1.size(); ++i) cout << ans1[i] << " ";
    cout << endl;
  } else {
    cout << ans.size() << endl;
    for (int i = 0; i < ans.size(); ++i) cout << ans[i] << " ";
    cout << endl;
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
const int N = (int)2e6 + 10;
int n;
bool is_prime[N];
vector<int> a;
vector<pair<int, int> > g[N];
set<int> was;
void build_prime() {
  for (int i = 2; i < N; ++i) is_prime[i] = true;
  for (int i = 2; i < N; ++i) {
    if (is_prime[i]) {
      int j = i + i;
      while (j < N) {
        is_prime[j] = false;
        j += i;
      }
    }
  }
}
int cnt1;
vector<int> ans1, ans;
int main() {
  build_prime();
  cin >> n;
  srand(time(NULL));
  for (int i = 0; i < n; ++i) {
    int x;
    cin >> x;
    if (x == 1) ans1.push_back(1);
    if (x == 1 || !was.count(x)) a.push_back(x);
    was.insert(x);
  }
  was.clear();
  for (int i = 0; i < a.size(); ++i)
    for (int j = 0; j < a.size(); ++j) {
      if (i == j) continue;
      if (is_prime[a[i] + a[j]]) {
        if (ans.size() == 0) {
          ans.push_back(a[i]);
          ans.push_back(a[j]);
        }
        if (a[i] != 1 && a[j] != 1 && is_prime[a[i] + 1] &&
            is_prime[a[j] + 1] && (ans1.size() != 0 || ans1.back() == 1)) {
          ans1.push_back(a[i]);
          ans1.push_back(a[j]);
        }
      }
    }
  if (ans.size() == 0) {
    for (int i = 0; i < a.size(); ++i) {
      ans.push_back(a[i]);
      break;
    }
  }
  if (ans1.size() != 0 && ans1.back() == 1) {
    for (int i = 0; i < a.size(); ++i) {
      if (is_prime[a[i] + 1] && a[i] != 1) {
        ans1.push_back(a[i]);
        break;
      }
    }
  }
  if (ans1.size() > ans.size()) {
    cout << ans1.size() << endl;
    for (int i = 0; i < ans1.size(); ++i) cout << ans1[i] << " ";
    cout << endl;
  } else {
    cout << ans.size() << endl;
    for (int i = 0; i < ans.size(); ++i) cout << ans[i] << " ";
    cout << endl;
  }
  return 0;
}

```