#include <bits/stdc++.h>
using namespace std;
bool prime[2000005];
void SieveOfEratosthenes(long long n) {
  memset(prime, true, sizeof(prime));
  for (long long p = 2; p * p <= 2000000; p++) {
    if (prime[p] == true) {
      for (long long i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  return;
}
int main() {
  SieveOfEratosthenes(2000000);
  long long t;
  t = 1;
  while (t--) {
    long long n;
    cin >> n;
    long long ar[n];
    for (long long i = 0; i < n; i++) cin >> ar[i];
    ;
    long long freq[1000001];
    memset(freq, 0, sizeof(freq));
    for (long long val : ar) freq[val]++;
    if (freq[1] < 2) {
      pair<long long, long long> res;
      long long flg = 0;
      for (long long i = 0; i < n; i++) {
        for (long long j = i + 1; j <= n - 1; j++) {
          if (prime[ar[i] + ar[j]]) {
            flg = 1;
            res = {ar[i], ar[j]};
            break;
          }
        }
        if (flg) break;
      }
      if (flg) {
        cout << 2 << "\n";
        cout << res.first << " " << res.second << "\n";
      } else {
        for (long long i = 0; i < n; i++) {
          if (prime[ar[i]]) {
            cout << "1\n" << ar[i] << "\n";
            flg = 1;
            break;
          }
        }
        if (flg == 0) {
          cout << "1\n";
          cout << ar[0] << "\n";
        }
      }
    } else {
      long long vl = -1;
      for (long long i = 0; i < n; i++) {
        if (ar[i] != 1 and prime[ar[i] + 1ll]) {
          vl = ar[i];
          break;
        }
      }
      if (vl != -1) {
        cout << (freq[1] + 1ll) << "\n";
      } else
        cout << freq[1] << "\n";
      for (long long i = 0; i < freq[1]; i++) cout << "1 ";
      if (vl != -1)
        cout << vl << "\n";
      else
        cout << "\n";
    }
  }
  return 0;
}

### 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;
bool prime[2000005];
void SieveOfEratosthenes(long long n) {
  memset(prime, true, sizeof(prime));
  for (long long p = 2; p * p <= 2000000; p++) {
    if (prime[p] == true) {
      for (long long i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
  return;
}
int main() {
  SieveOfEratosthenes(2000000);
  long long t;
  t = 1;
  while (t--) {
    long long n;
    cin >> n;
    long long ar[n];
    for (long long i = 0; i < n; i++) cin >> ar[i];
    ;
    long long freq[1000001];
    memset(freq, 0, sizeof(freq));
    for (long long val : ar) freq[val]++;
    if (freq[1] < 2) {
      pair<long long, long long> res;
      long long flg = 0;
      for (long long i = 0; i < n; i++) {
        for (long long j = i + 1; j <= n - 1; j++) {
          if (prime[ar[i] + ar[j]]) {
            flg = 1;
            res = {ar[i], ar[j]};
            break;
          }
        }
        if (flg) break;
      }
      if (flg) {
        cout << 2 << "\n";
        cout << res.first << " " << res.second << "\n";
      } else {
        for (long long i = 0; i < n; i++) {
          if (prime[ar[i]]) {
            cout << "1\n" << ar[i] << "\n";
            flg = 1;
            break;
          }
        }
        if (flg == 0) {
          cout << "1\n";
          cout << ar[0] << "\n";
        }
      }
    } else {
      long long vl = -1;
      for (long long i = 0; i < n; i++) {
        if (ar[i] != 1 and prime[ar[i] + 1ll]) {
          vl = ar[i];
          break;
        }
      }
      if (vl != -1) {
        cout << (freq[1] + 1ll) << "\n";
      } else
        cout << freq[1] << "\n";
      for (long long i = 0; i < freq[1]; i++) cout << "1 ";
      if (vl != -1)
        cout << vl << "\n";
      else
        cout << "\n";
    }
  }
  return 0;
}

```