#include <bits/stdc++.h>
const int N = 2000001;
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  int a[n];
  int one = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    one += a[i] == 1;
  }
  bool pr[N];
  memset(pr, true, sizeof pr);
  pr[0] = pr[1] = false;
  for (int i = 2; i < N; i++) {
    if (pr[i]) {
      for (int j = i + i; j < N; j += i) {
        pr[j] = false;
      }
    }
  }
  if (one >= 2) {
    for (int e : a) {
      if (e % 2 == 0 && pr[e + 1]) {
        cout << one + 1 << '\n';
        while (one--) {
          cout << "1 ";
        }
        cout << e;
        return 0;
      }
    }
    cout << one << '\n';
    while (one--) {
      cout << "1 ";
    }
    return 0;
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (pr[a[i] + a[j]]) {
        cout << "2\n" << a[i] << ' ' << a[j];
        return 0;
      }
    }
  }
  cout << "1\n" << a[0];
}

### Prompt

In cpp, your task is to solve 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>
const int N = 2000001;
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  int a[n];
  int one = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    one += a[i] == 1;
  }
  bool pr[N];
  memset(pr, true, sizeof pr);
  pr[0] = pr[1] = false;
  for (int i = 2; i < N; i++) {
    if (pr[i]) {
      for (int j = i + i; j < N; j += i) {
        pr[j] = false;
      }
    }
  }
  if (one >= 2) {
    for (int e : a) {
      if (e % 2 == 0 && pr[e + 1]) {
        cout << one + 1 << '\n';
        while (one--) {
          cout << "1 ";
        }
        cout << e;
        return 0;
      }
    }
    cout << one << '\n';
    while (one--) {
      cout << "1 ";
    }
    return 0;
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (pr[a[i] + a[j]]) {
        cout << "2\n" << a[i] << ' ' << a[j];
        return 0;
      }
    }
  }
  cout << "1\n" << a[0];
}

```