#include <bits/stdc++.h>
using namespace std;
int lp[2000020];
map<int, int> z;
vector<int> pr;
int n, k1, f = 0;
int a[1000010];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 2; i < 2000010; ++i) {
    if (lp[i] == 0) {
      lp[i] = i;
      pr.push_back(i);
    }
    for (int j = 0; j < pr.size() && pr[j] <= lp[i] && i * pr[j] < 2000010; j++)
      lp[i * pr[j]] = pr[j];
  }
  for (int i = 0; i < pr.size(); i++) z.insert({pr[i], 0});
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 1) k1++;
    if (a[i] != 1 && lp[a[i] + 1] == a[i] + 1) f = a[i];
  }
  if (k1 + min(f, 1) >= 2) {
    cout << min(f, 1) + k1 << endl;
    if (f != 0) cout << f << ' ';
    for (int i = 0; i < k1; i++) cout << 1 << ' ';
    return 0;
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      if (lp[a[i] + a[j]] == a[i] + a[j]) {
        cout << 2 << endl;
        cout << a[i] << ' ' << a[j] << endl;
        return 0;
      }
    }
  cout << 1 << endl << a[0] << endl;
}

### Prompt

Create a solution in CPP for 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;
int lp[2000020];
map<int, int> z;
vector<int> pr;
int n, k1, f = 0;
int a[1000010];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 2; i < 2000010; ++i) {
    if (lp[i] == 0) {
      lp[i] = i;
      pr.push_back(i);
    }
    for (int j = 0; j < pr.size() && pr[j] <= lp[i] && i * pr[j] < 2000010; j++)
      lp[i * pr[j]] = pr[j];
  }
  for (int i = 0; i < pr.size(); i++) z.insert({pr[i], 0});
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 1) k1++;
    if (a[i] != 1 && lp[a[i] + 1] == a[i] + 1) f = a[i];
  }
  if (k1 + min(f, 1) >= 2) {
    cout << min(f, 1) + k1 << endl;
    if (f != 0) cout << f << ' ';
    for (int i = 0; i < k1; i++) cout << 1 << ' ';
    return 0;
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      if (lp[a[i] + a[j]] == a[i] + a[j]) {
        cout << 2 << endl;
        cout << a[i] << ' ' << a[j] << endl;
        return 0;
      }
    }
  cout << 1 << endl << a[0] << endl;
}

```