#include <bits/stdc++.h>
using namespace std;
int dx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dy[] = {-1, 0, 1, -1, 1, -1, 0, 1};
void fast() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
vector<bool> temp(2e6 + 5, false);
void sieve(long long n) {
  vector<bool> isPrime(n + 1, true);
  isPrime[0] = isPrime[1] = 0;
  for (long long i = 2; i * i <= n; ++i) {
    if (isPrime[i]) {
      for (int j = i * 2; j <= n; j += i) isPrime[j] = 0;
    }
  }
  int cnt = 0;
  for (long long i = 0; i < (int)isPrime.size(); ++i)
    if (isPrime[i]) temp[i] = true;
}
int main() {
  fast();
  sieve(2e6 + 5);
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  vector<long long> v[(int)1e3 + 10];
  sort(arr, arr + n);
  for (auto x : arr) {
    for (auto &vec : v) {
      bool flag = 0;
      for (auto &z : vec) {
        if (!temp[z + x]) {
          flag = 1;
          break;
        }
      }
      if (!flag) {
        vec.push_back(x);
        break;
      }
    }
  }
  long long mx = 0, idx = -1;
  for (int i = 0; i < 1e3; i++) {
    if (mx < (int)v[i].size()) mx = (int)v[i].size(), idx = i;
  }
  cout << (int)v[idx].size() << "\n";
  for (auto x : v[idx]) cout << x << ' ';
}

### Prompt

Your task is to create 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 dx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dy[] = {-1, 0, 1, -1, 1, -1, 0, 1};
void fast() {
  std::ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
}
vector<bool> temp(2e6 + 5, false);
void sieve(long long n) {
  vector<bool> isPrime(n + 1, true);
  isPrime[0] = isPrime[1] = 0;
  for (long long i = 2; i * i <= n; ++i) {
    if (isPrime[i]) {
      for (int j = i * 2; j <= n; j += i) isPrime[j] = 0;
    }
  }
  int cnt = 0;
  for (long long i = 0; i < (int)isPrime.size(); ++i)
    if (isPrime[i]) temp[i] = true;
}
int main() {
  fast();
  sieve(2e6 + 5);
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  vector<long long> v[(int)1e3 + 10];
  sort(arr, arr + n);
  for (auto x : arr) {
    for (auto &vec : v) {
      bool flag = 0;
      for (auto &z : vec) {
        if (!temp[z + x]) {
          flag = 1;
          break;
        }
      }
      if (!flag) {
        vec.push_back(x);
        break;
      }
    }
  }
  long long mx = 0, idx = -1;
  for (int i = 0; i < 1e3; i++) {
    if (mx < (int)v[i].size()) mx = (int)v[i].size(), idx = i;
  }
  cout << (int)v[idx].size() << "\n";
  for (auto x : v[idx]) cout << x << ' ';
}

```