#include <bits/stdc++.h>
using namespace std;
const int N = 1 << 21;
int main() {
  vector<bool> is_prime(N, true);
  is_prime[0] = is_prime[1] = false;
  for (int i = 2; i * i < N; i++)
    if (is_prime[i])
      for (int j = i * i; j < N; j += i) is_prime[j] = false;
  int n, one = 0;
  scanf("%d", &n);
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) one++;
  }
  vector<int> ans(one, 1);
  if (one > 0) {
    for (int i = 0; i < n; i++)
      if (a[i] != 1 && is_prime[a[i] + 1]) {
        ans.push_back(a[i]);
        break;
      }
  }
  if (ans.size() < 2) {
    ans.clear();
    for (int i = 0; ans.size() < 2 && i < n; i++)
      for (int j = i + 1; ans.size() < 2 && j < n; j++)
        if (is_prime[a[i] + a[j]]) {
          ans.push_back(a[i]);
          ans.push_back(a[j]);
        }
  }
  if (ans.empty()) ans.push_back(a[0]);
  printf("%lu\n", ans.size());
  for (int i = ans.size() - 1; i >= 0; i--)
    printf(i == 0 ? "%d\n" : "%d ", ans[i]);
}

### 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;
const int N = 1 << 21;
int main() {
  vector<bool> is_prime(N, true);
  is_prime[0] = is_prime[1] = false;
  for (int i = 2; i * i < N; i++)
    if (is_prime[i])
      for (int j = i * i; j < N; j += i) is_prime[j] = false;
  int n, one = 0;
  scanf("%d", &n);
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) one++;
  }
  vector<int> ans(one, 1);
  if (one > 0) {
    for (int i = 0; i < n; i++)
      if (a[i] != 1 && is_prime[a[i] + 1]) {
        ans.push_back(a[i]);
        break;
      }
  }
  if (ans.size() < 2) {
    ans.clear();
    for (int i = 0; ans.size() < 2 && i < n; i++)
      for (int j = i + 1; ans.size() < 2 && j < n; j++)
        if (is_prime[a[i] + a[j]]) {
          ans.push_back(a[i]);
          ans.push_back(a[j]);
        }
  }
  if (ans.empty()) ans.push_back(a[0]);
  printf("%lu\n", ans.size());
  for (int i = ans.size() - 1; i >= 0; i--)
    printf(i == 0 ? "%d\n" : "%d ", ans[i]);
}

```