#include <bits/stdc++.h>
using namespace std;
const int maxx = 2 * 1e6 + 7, mod = 1e9 + 7;
int n, a[maxx], prime[maxx], vis[maxx];
void getprime(int x) {
  for (int i = 2; i <= x; i++) {
    if (!vis[i]) prime[++prime[0]] = i;
    for (int j = 1; j <= prime[0] && i * prime[j] <= x; j++) {
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
int main() {
  getprime(2000000);
  int one = 0, ans = 0, two = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == 1) one++;
    if (a[i] % 2 == 0 && !vis[a[i] + 1]) two = a[i];
  }
  ans = one + (two != 0);
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (!vis[a[i] + a[j]] && ans <= 2) {
        cout << 2 << '\n' << a[i] << " " << a[j] << endl;
        goto next;
      }
    }
  }
  if (ans) {
    cout << ans << endl;
    for (int i = 1; i <= one; i++) cout << 1 << " ";
    if (two)
      cout << two << endl;
    else
      cout << endl;
  } else
    cout << 1 << '\n' << a[1] << endl;
next:;
}

### Prompt

Generate 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 int maxx = 2 * 1e6 + 7, mod = 1e9 + 7;
int n, a[maxx], prime[maxx], vis[maxx];
void getprime(int x) {
  for (int i = 2; i <= x; i++) {
    if (!vis[i]) prime[++prime[0]] = i;
    for (int j = 1; j <= prime[0] && i * prime[j] <= x; j++) {
      vis[i * prime[j]] = 1;
      if (i % prime[j] == 0) break;
    }
  }
}
int main() {
  getprime(2000000);
  int one = 0, ans = 0, two = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == 1) one++;
    if (a[i] % 2 == 0 && !vis[a[i] + 1]) two = a[i];
  }
  ans = one + (two != 0);
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (!vis[a[i] + a[j]] && ans <= 2) {
        cout << 2 << '\n' << a[i] << " " << a[j] << endl;
        goto next;
      }
    }
  }
  if (ans) {
    cout << ans << endl;
    for (int i = 1; i <= one; i++) cout << 1 << " ";
    if (two)
      cout << two << endl;
    else
      cout << endl;
  } else
    cout << 1 << '\n' << a[1] << endl;
next:;
}

```