#include <bits/stdc++.h>
using namespace std;
const int N = 1000;
int n, a[N], b[N][N], p[N], s = 0;
vector<bool> prime(2000001, true);
int main() {
  for (int i = 2; i < 2000; i++)
    if (prime[i])
      for (int j = i * i; j < 2000001; j += i) prime[j] = false;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i != j && !prime[a[i] + a[j]]) b[i][j] = 1, s++;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (b[i][j]) p[i]++;
  list<int> pq;
  for (int i = 0; i < n; i++) pq.push_back(i);
  while (s) {
    int best = pq.front();
    for (int it : pq)
      if (p[it] > p[best]) best = it;
    s -= 2 * p[best];
    for (int j = 0; j < n; j++)
      if (b[best][j] == 1) p[j]--;
    for (auto it = pq.begin(); it != pq.end(); it++)
      if (*it == best) {
        pq.erase(it);
        break;
      }
  }
  cout << ((int)pq.size()) << endl;
  for (int it : pq) cout << a[it] << ' ';
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution 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;
const int N = 1000;
int n, a[N], b[N][N], p[N], s = 0;
vector<bool> prime(2000001, true);
int main() {
  for (int i = 2; i < 2000; i++)
    if (prime[i])
      for (int j = i * i; j < 2000001; j += i) prime[j] = false;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i != j && !prime[a[i] + a[j]]) b[i][j] = 1, s++;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (b[i][j]) p[i]++;
  list<int> pq;
  for (int i = 0; i < n; i++) pq.push_back(i);
  while (s) {
    int best = pq.front();
    for (int it : pq)
      if (p[it] > p[best]) best = it;
    s -= 2 * p[best];
    for (int j = 0; j < n; j++)
      if (b[best][j] == 1) p[j]--;
    for (auto it = pq.begin(); it != pq.end(); it++)
      if (*it == best) {
        pq.erase(it);
        break;
      }
  }
  cout << ((int)pq.size()) << endl;
  for (int it : pq) cout << a[it] << ' ';
  return 0;
}

```