#include <bits/stdc++.h>
using namespace std;
const long long int N = 1000000 + 5;
long long int arr[N];
long long int n;
long long int prime[2 * N + 5];
void sieve() {
  memset(prime, 1, sizeof(prime));
  for (long long int i = 2; i < 2 * N; i++)
    for (long long int j = 2; j * i < 2 * N; j++) prime[i * j] = 0;
}
void solve() {
  cin >> n;
  long long int cnt[N] = {0};
  for (long long int i = 1; i <= n; i++) {
    cin >> arr[i];
    cnt[arr[i]]++;
  }
  sieve();
  long long int ones = cnt[1];
  if (ones > 0) {
    for (long long int i = 3; i <= N; i++) {
      if (prime[i] && cnt[i - 1]) {
        cout << ones + 1 << endl;
        for (long long int j = 1; j <= ones; j++) cout << 1 << " ";
        cout << i - 1 << endl;
        return;
      }
    }
    if (ones > 1) {
      cout << ones << endl;
      for (long long int i = 1; i <= ones; i++) cout << 1 << " ";
      return;
    }
  }
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = i + 1; j <= n; j++) {
      if (prime[arr[i] + arr[j]]) {
        cout << 2 << endl;
        cout << arr[i] << " " << arr[j] << endl;
        return;
      }
    }
  }
  cout << 1 << endl;
  cout << arr[1] << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### 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;
const long long int N = 1000000 + 5;
long long int arr[N];
long long int n;
long long int prime[2 * N + 5];
void sieve() {
  memset(prime, 1, sizeof(prime));
  for (long long int i = 2; i < 2 * N; i++)
    for (long long int j = 2; j * i < 2 * N; j++) prime[i * j] = 0;
}
void solve() {
  cin >> n;
  long long int cnt[N] = {0};
  for (long long int i = 1; i <= n; i++) {
    cin >> arr[i];
    cnt[arr[i]]++;
  }
  sieve();
  long long int ones = cnt[1];
  if (ones > 0) {
    for (long long int i = 3; i <= N; i++) {
      if (prime[i] && cnt[i - 1]) {
        cout << ones + 1 << endl;
        for (long long int j = 1; j <= ones; j++) cout << 1 << " ";
        cout << i - 1 << endl;
        return;
      }
    }
    if (ones > 1) {
      cout << ones << endl;
      for (long long int i = 1; i <= ones; i++) cout << 1 << " ";
      return;
    }
  }
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = i + 1; j <= n; j++) {
      if (prime[arr[i] + arr[j]]) {
        cout << 2 << endl;
        cout << arr[i] << " " << arr[j] << endl;
        return;
      }
    }
  }
  cout << 1 << endl;
  cout << arr[1] << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```