#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long mod1 = 998244353;
const int N = 2e6 + 12;
bool p[N];
long long a[2000], n, c = 0;
vector<int> ans;
void show() {
  cout << (long long)ans.size() << "\n";
  for (auto it : ans) cout << it << " ";
  return;
}
void solve() {
  for (int i = 0; i < N; i++) p[i] = 1;
  p[0] = p[1] = 0;
  for (long long i = 2; i < N; i++)
    if (p[i]) {
      for (long long j = i * i; j < N; j += i) p[j] = 0;
    }
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  for (int i = 0; i < n; i++) c += (a[i] == 1);
  if (c > 1) {
    for (int i = 0; i < c; i++) ans.push_back(1);
    for (int i = c; i < n; i++)
      if (p[a[i] + 1]) {
        ans.push_back(a[i]);
        break;
      }
    show();
    return;
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (p[a[i] + a[j]]) {
        ans.push_back(a[i]);
        ans.push_back(a[j]);
        show();
        return;
      }
    }
  }
  ans.push_back(a[0]);
  show();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Your task is to create 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;
long long mod = 1e9 + 7;
long long mod1 = 998244353;
const int N = 2e6 + 12;
bool p[N];
long long a[2000], n, c = 0;
vector<int> ans;
void show() {
  cout << (long long)ans.size() << "\n";
  for (auto it : ans) cout << it << " ";
  return;
}
void solve() {
  for (int i = 0; i < N; i++) p[i] = 1;
  p[0] = p[1] = 0;
  for (long long i = 2; i < N; i++)
    if (p[i]) {
      for (long long j = i * i; j < N; j += i) p[j] = 0;
    }
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  for (int i = 0; i < n; i++) c += (a[i] == 1);
  if (c > 1) {
    for (int i = 0; i < c; i++) ans.push_back(1);
    for (int i = c; i < n; i++)
      if (p[a[i] + 1]) {
        ans.push_back(a[i]);
        break;
      }
    show();
    return;
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (p[a[i] + a[j]]) {
        ans.push_back(a[i]);
        ans.push_back(a[j]);
        show();
        return;
      }
    }
  }
  ans.push_back(a[0]);
  show();
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```