#include <bits/stdc++.h>
using namespace std;
int spf[2000005];
void sieve() {
  spf[1] = 1;
  for (int i = 2; i <= 2e6; i++) spf[i] = i;
  for (int i = 4; i <= 2e6; i += 2) spf[i] = 2;
  for (int i = 3; i * i <= 2e6; i += 2) {
    if (spf[i] == i) {
      for (int j = i * i; j <= 2e6; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
int main() {
  sieve();
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  sort(arr, arr + n);
  vector<int> ans;
  if (arr[0] == 1) {
    for (int i = 0; i < n; i++)
      if (arr[i] == 1) ans.push_back(1);
    for (int j = 0; j < n; j++) {
      if (arr[j] != 1 && spf[arr[j] + 1] == arr[j] + 1) {
        ans.push_back(arr[j]);
        break;
      }
    }
  }
  if (ans.size() <= 1) {
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        if (spf[arr[i] + arr[j]] == arr[i] + arr[j])
          return cout << 2 << endl << arr[i] << ' ' << arr[j], 0;
      }
    }
  }
  if (ans.size() == 0) return cout << 1 << endl << arr[0], 0;
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << ' ';
  return 0;
}

### Prompt

Please create a solution in cpp to 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 spf[2000005];
void sieve() {
  spf[1] = 1;
  for (int i = 2; i <= 2e6; i++) spf[i] = i;
  for (int i = 4; i <= 2e6; i += 2) spf[i] = 2;
  for (int i = 3; i * i <= 2e6; i += 2) {
    if (spf[i] == i) {
      for (int j = i * i; j <= 2e6; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
int main() {
  sieve();
  int n;
  cin >> n;
  int arr[n];
  for (int i = 0; i < n; i++) cin >> arr[i];
  sort(arr, arr + n);
  vector<int> ans;
  if (arr[0] == 1) {
    for (int i = 0; i < n; i++)
      if (arr[i] == 1) ans.push_back(1);
    for (int j = 0; j < n; j++) {
      if (arr[j] != 1 && spf[arr[j] + 1] == arr[j] + 1) {
        ans.push_back(arr[j]);
        break;
      }
    }
  }
  if (ans.size() <= 1) {
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        if (spf[arr[i] + arr[j]] == arr[i] + arr[j])
          return cout << 2 << endl << arr[i] << ' ' << arr[j], 0;
      }
    }
  }
  if (ans.size() == 0) return cout << 1 << endl << arr[0], 0;
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << ' ';
  return 0;
}

```