#include <bits/stdc++.h>
using namespace std;
const long long N = 2e6 + 5;
long long n;
long long a[N], f[N];
long long sieve[N];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  fill(sieve + 2, sieve + N, 1);
  for (long long i = 2; i < N; i++) {
    for (long long j = 2; i * j < N; j++) sieve[i * j] = 0;
  }
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    f[a[i]]++;
  }
  long long ans = 0;
  long long ones = f[1];
  if (ones > 0) {
    for (long long i = 3; i < N; i++) {
      if (sieve[i] && f[i - 1]) {
        cout << ones + 1 << "\n";
        for (long long j = 1; j <= ones; j++) cout << 1 << " ";
        cout << i - 1;
        return 0;
      }
    }
    if (ones > 1) {
      cout << ones << "\n";
      for (long long j = 1; j <= ones; j++) cout << 1 << " ";
      return 0;
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = i + 1; j <= n; j++) {
      if (sieve[a[i] + a[j]]) {
        cout << 2 << "\n";
        cout << a[i] << " " << a[j];
        return 0;
      }
    }
  }
  cout << 1 << "\n";
  cout << a[1];
  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;
const long long N = 2e6 + 5;
long long n;
long long a[N], f[N];
long long sieve[N];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  fill(sieve + 2, sieve + N, 1);
  for (long long i = 2; i < N; i++) {
    for (long long j = 2; i * j < N; j++) sieve[i * j] = 0;
  }
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    f[a[i]]++;
  }
  long long ans = 0;
  long long ones = f[1];
  if (ones > 0) {
    for (long long i = 3; i < N; i++) {
      if (sieve[i] && f[i - 1]) {
        cout << ones + 1 << "\n";
        for (long long j = 1; j <= ones; j++) cout << 1 << " ";
        cout << i - 1;
        return 0;
      }
    }
    if (ones > 1) {
      cout << ones << "\n";
      for (long long j = 1; j <= ones; j++) cout << 1 << " ";
      return 0;
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = i + 1; j <= n; j++) {
      if (sieve[a[i] + a[j]]) {
        cout << 2 << "\n";
        cout << a[i] << " " << a[j];
        return 0;
      }
    }
  }
  cout << 1 << "\n";
  cout << a[1];
  return 0;
}

```