#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 1;
const int INF = 1e9 + 9;
const int B = 1e9 + 7;
int n, m, k;
int a[N], c[N];
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  int cnt1 = 0;
  for (int i = int(1); i <= int(n); ++i) {
    cin >> a[i];
    cnt1 += (a[i] == 1);
  }
  for (int i = int(2); i <= int(N - 1); ++i)
    if (!c[i])
      for (long long j = 1ll * i * i; j < N; j += i) c[j] = 1;
  for (int i = int(1); i <= int(n); ++i) {
    if (a[i] > 1 && !c[a[i] + 1] && cnt1 > 0) {
      cout << 1 + cnt1 << "\n";
      for (int j = int(1); j <= int(cnt1); ++j) cout << "1 ";
      cout << a[i] << " ";
      cout << "\n";
      return 0;
    }
  }
  if (cnt1 >= 2) {
    cout << cnt1 << "\n";
    for (int j = int(1); j <= int(cnt1); ++j) cout << "1 ";
    cout << "\n";
    return 0;
  }
  for (int i = int(1); i <= int(n); ++i) {
    for (int j = int(1); j <= int(i - 1); ++j) {
      if (!c[a[i] + a[j]]) {
        cout << "2\n";
        cout << a[i] << " " << a[j] << "\n";
        return 0;
      }
    }
  }
  cout << "1\n" << a[1] << "\n";
  return 0;
}

### Prompt

Create a solution in CPP for 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 = 2e6 + 1;
const int INF = 1e9 + 9;
const int B = 1e9 + 7;
int n, m, k;
int a[N], c[N];
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  int cnt1 = 0;
  for (int i = int(1); i <= int(n); ++i) {
    cin >> a[i];
    cnt1 += (a[i] == 1);
  }
  for (int i = int(2); i <= int(N - 1); ++i)
    if (!c[i])
      for (long long j = 1ll * i * i; j < N; j += i) c[j] = 1;
  for (int i = int(1); i <= int(n); ++i) {
    if (a[i] > 1 && !c[a[i] + 1] && cnt1 > 0) {
      cout << 1 + cnt1 << "\n";
      for (int j = int(1); j <= int(cnt1); ++j) cout << "1 ";
      cout << a[i] << " ";
      cout << "\n";
      return 0;
    }
  }
  if (cnt1 >= 2) {
    cout << cnt1 << "\n";
    for (int j = int(1); j <= int(cnt1); ++j) cout << "1 ";
    cout << "\n";
    return 0;
  }
  for (int i = int(1); i <= int(n); ++i) {
    for (int j = int(1); j <= int(i - 1); ++j) {
      if (!c[a[i] + a[j]]) {
        cout << "2\n";
        cout << a[i] << " " << a[j] << "\n";
        return 0;
      }
    }
  }
  cout << "1\n" << a[1] << "\n";
  return 0;
}

```