#include <bits/stdc++.h>
using namespace std;
bool prime[2000001];
int main() {
  memset(prime, true, sizeof(prime));
  int temp2, temp;
  for (temp = 2; temp <= 2000001; temp++) {
    if (prime[temp]) {
      for (temp2 = temp * 2; temp2 <= 2000001; temp2 += temp) {
        prime[temp2] = false;
      }
    }
  }
  int n;
  cin >> n;
  int freq1 = 0, numf = 0;
  int a[n];
  vector<int> even;
  vector<int> odd;
  for (temp = 0; temp < n; temp++) {
    cin >> a[temp];
    if (a[temp] % 2 == 0)
      even.push_back(a[temp]);
    else
      odd.push_back(a[temp]);
    if (a[temp] == 1)
      freq1++;
    else if (prime[a[temp] + 1])
      numf = a[temp];
  }
  int cnt = freq1;
  if (numf) cnt++;
  if (cnt >= 2) {
    cout << cnt << "\n";
    for (temp = 0; temp < freq1; temp++) {
      cout << 1;
      if (temp != freq1 - 1) {
        cout << " ";
      }
    }
    if (numf) cout << " " << numf;
    cout << "\n";
    return 0;
  }
  for (temp = 0; temp < n; temp++) {
    for (temp2 = 0; temp2 < even.size(); temp2++) {
      if (prime[a[temp] + even[temp2]]) {
        cout << 2 << "\n";
        cout << a[temp] << " " << even[temp2] << "\n";
        return 0;
      }
    }
  }
  cout << 1 << "\n";
  cout << a[0] << "\n";
}

### 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;
bool prime[2000001];
int main() {
  memset(prime, true, sizeof(prime));
  int temp2, temp;
  for (temp = 2; temp <= 2000001; temp++) {
    if (prime[temp]) {
      for (temp2 = temp * 2; temp2 <= 2000001; temp2 += temp) {
        prime[temp2] = false;
      }
    }
  }
  int n;
  cin >> n;
  int freq1 = 0, numf = 0;
  int a[n];
  vector<int> even;
  vector<int> odd;
  for (temp = 0; temp < n; temp++) {
    cin >> a[temp];
    if (a[temp] % 2 == 0)
      even.push_back(a[temp]);
    else
      odd.push_back(a[temp]);
    if (a[temp] == 1)
      freq1++;
    else if (prime[a[temp] + 1])
      numf = a[temp];
  }
  int cnt = freq1;
  if (numf) cnt++;
  if (cnt >= 2) {
    cout << cnt << "\n";
    for (temp = 0; temp < freq1; temp++) {
      cout << 1;
      if (temp != freq1 - 1) {
        cout << " ";
      }
    }
    if (numf) cout << " " << numf;
    cout << "\n";
    return 0;
  }
  for (temp = 0; temp < n; temp++) {
    for (temp2 = 0; temp2 < even.size(); temp2++) {
      if (prime[a[temp] + even[temp2]]) {
        cout << 2 << "\n";
        cout << a[temp] << " " << even[temp2] << "\n";
        return 0;
      }
    }
  }
  cout << 1 << "\n";
  cout << a[0] << "\n";
}

```