#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 5;
int is[N];
int a[1005];
void init() {
  memset(is, 1, sizeof(is));
  is[1] = is[0] = 0;
  for (int i = 2; i < N; i++) {
    if (is[i] == 0) continue;
    for (int j = 2 * i; j < N; j += i) is[j] = 0;
  }
}
vector<int> v;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  init();
  int n;
  cin >> n;
  int ans = 0;
  int cnt1 = 0;
  int other = 0, oans = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == 1) cnt1++;
    if (is[a[i] + 1] && a[i] != 1) {
      other++;
      oans = a[i];
    }
    if (true) {
      v.clear();
      v.push_back(a[i]);
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++)
      if (is[a[i] + a[j]]) {
        v.clear();
        v.push_back(a[i]);
        v.push_back(a[j]);
      }
  if (cnt1 > 0) {
    if (cnt1 == 1 && other == 0)
      ;
    else {
      ans = cnt1 + ((other >= 1) ? 1 : 0);
      v.clear();
      for (int i = 1; i <= cnt1; i++) v.push_back(1);
      if (other >= 1) v.push_back(oans);
    }
  }
  cout << v.size() << endl;
  for (int i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 N = 2e6 + 5;
int is[N];
int a[1005];
void init() {
  memset(is, 1, sizeof(is));
  is[1] = is[0] = 0;
  for (int i = 2; i < N; i++) {
    if (is[i] == 0) continue;
    for (int j = 2 * i; j < N; j += i) is[j] = 0;
  }
}
vector<int> v;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  init();
  int n;
  cin >> n;
  int ans = 0;
  int cnt1 = 0;
  int other = 0, oans = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == 1) cnt1++;
    if (is[a[i] + 1] && a[i] != 1) {
      other++;
      oans = a[i];
    }
    if (true) {
      v.clear();
      v.push_back(a[i]);
    }
  }
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++)
      if (is[a[i] + a[j]]) {
        v.clear();
        v.push_back(a[i]);
        v.push_back(a[j]);
      }
  if (cnt1 > 0) {
    if (cnt1 == 1 && other == 0)
      ;
    else {
      ans = cnt1 + ((other >= 1) ? 1 : 0);
      v.clear();
      for (int i = 1; i <= cnt1; i++) v.push_back(1);
      if (other >= 1) v.push_back(oans);
    }
  }
  cout << v.size() << endl;
  for (int i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << endl;
  return 0;
}

```