#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e6 + 100;
int a[maxn];
bool p[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  p[0] = p[1] = true;
  for (int i = 2; i < maxn; i++)
    for (int j = 2; i * j < maxn; j++) p[i * j] = true;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  int cnt = 0;
  for (int i = 0; i < n; i++)
    if (a[i] == 1) cnt++;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < i; j++)
      if (a[i] > 1 && a[j] > 1 && !p[a[i] + 1] && !p[a[j] + 1] &&
          !p[a[i] + a[j]]) {
        cout << cnt + 2 << endl;
        for (int k = 0; k < cnt; k++) cout << "1 ";
        cout << a[i] << " " << a[j] << endl;
        return 0;
      }
  if (cnt >= 1) {
    for (int i = 0; i < n; i++)
      if (a[i] > 1 && !p[a[i] + 1]) {
        cout << cnt + 1 << endl;
        for (int j = 0; j < cnt; j++) cout << "1 ";
        cout << a[i] << endl;
        return 0;
      }
    if (cnt > 1) {
      cout << cnt << endl;
      for (int i = 0; i < cnt; i++) cout << "1 ";
      cout << endl;
      return 0;
    }
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < i; j++)
      if (a[i] > 1 && a[j] > 1 && !p[a[i] + a[j]]) {
        cout << 2 << endl;
        cout << a[i] << " " << a[j] << endl;
        return 0;
      }
  cout << 1 << endl;
  cout << a[0] << endl;
  return 0;
}

### Prompt

Develop a solution in CPP 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 int maxn = 2e6 + 100;
int a[maxn];
bool p[maxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  p[0] = p[1] = true;
  for (int i = 2; i < maxn; i++)
    for (int j = 2; i * j < maxn; j++) p[i * j] = true;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  int cnt = 0;
  for (int i = 0; i < n; i++)
    if (a[i] == 1) cnt++;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < i; j++)
      if (a[i] > 1 && a[j] > 1 && !p[a[i] + 1] && !p[a[j] + 1] &&
          !p[a[i] + a[j]]) {
        cout << cnt + 2 << endl;
        for (int k = 0; k < cnt; k++) cout << "1 ";
        cout << a[i] << " " << a[j] << endl;
        return 0;
      }
  if (cnt >= 1) {
    for (int i = 0; i < n; i++)
      if (a[i] > 1 && !p[a[i] + 1]) {
        cout << cnt + 1 << endl;
        for (int j = 0; j < cnt; j++) cout << "1 ";
        cout << a[i] << endl;
        return 0;
      }
    if (cnt > 1) {
      cout << cnt << endl;
      for (int i = 0; i < cnt; i++) cout << "1 ";
      cout << endl;
      return 0;
    }
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < i; j++)
      if (a[i] > 1 && a[j] > 1 && !p[a[i] + a[j]]) {
        cout << 2 << endl;
        cout << a[i] << " " << a[j] << endl;
        return 0;
      }
  cout << 1 << endl;
  cout << a[0] << endl;
  return 0;
}

```