#include <bits/stdc++.h>
using namespace std;
int Set(int N, int pos) { return N = N | (1 << pos); }
int reset(int N, int pos) { return N = N & ~(1 << pos); }
bool check(int N, int pos) { return (bool)(N & (1 << pos)); }
bool Notprime[4000005];
int main() {
  Notprime[1] = true;
  for (int i = 2; (i * i) < 4000005; i++) {
    if (Notprime[i] == true) continue;
    for (int j = i * 2; j < 4000005; j += i) Notprime[j] = true;
  }
  int n;
  scanf("%d", &n);
  int a[n + 1];
  int on = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  vector<int> ans;
  bool take = true;
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1)
      ans.push_back(1);
    else if (take && Notprime[a[i] + 1] == false) {
      ans.push_back(a[i]);
      take = false;
    }
  }
  if (ans.size() == 0) ans.push_back(a[1]);
  if (ans.size() < 2) {
    for (int i = 1; i <= n; i++) {
      for (int j = i + 1; j <= n; j++) {
        if (Notprime[a[i] + a[j]] == false) {
          ans.clear();
          ans.push_back(a[i]);
          ans.push_back(a[j]);
          n = -1;
          break;
        }
      }
    }
  }
  cout << ans.size() << endl;
  for (int i : ans) cout << i << " ";
  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;
int Set(int N, int pos) { return N = N | (1 << pos); }
int reset(int N, int pos) { return N = N & ~(1 << pos); }
bool check(int N, int pos) { return (bool)(N & (1 << pos)); }
bool Notprime[4000005];
int main() {
  Notprime[1] = true;
  for (int i = 2; (i * i) < 4000005; i++) {
    if (Notprime[i] == true) continue;
    for (int j = i * 2; j < 4000005; j += i) Notprime[j] = true;
  }
  int n;
  scanf("%d", &n);
  int a[n + 1];
  int on = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  vector<int> ans;
  bool take = true;
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1)
      ans.push_back(1);
    else if (take && Notprime[a[i] + 1] == false) {
      ans.push_back(a[i]);
      take = false;
    }
  }
  if (ans.size() == 0) ans.push_back(a[1]);
  if (ans.size() < 2) {
    for (int i = 1; i <= n; i++) {
      for (int j = i + 1; j <= n; j++) {
        if (Notprime[a[i] + a[j]] == false) {
          ans.clear();
          ans.push_back(a[i]);
          ans.push_back(a[j]);
          n = -1;
          break;
        }
      }
    }
  }
  cout << ans.size() << endl;
  for (int i : ans) cout << i << " ";
  return 0;
}

```