#include <bits/stdc++.h>
using namespace std;
const int N = 1010;
const int K = int(2e6) + 10;
const int MOD = int(1e9) + 7;
const int INF = int(2e9) + 5;
const long long INF64 = 1e18;
int a[N];
int sieve[K];
void calc_sieve(int n) {
  sieve[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (sieve[i]) continue;
    for (int u = 2 * i; u <= n; u += i) {
      sieve[u] = i;
    }
  }
}
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  calc_sieve(2 * a[n]);
  if (a[1] == 1 && a[2] == 1) {
    vector<int> ans;
    ans.push_back(1);
    for (int i = 2; i <= n; i++) {
      if (a[i] == 1) {
        ans.push_back(1);
        continue;
      }
      if (a[i] % 2 == 0 && sieve[1 + a[i]] == 0) {
        ans.push_back(a[i]);
        break;
      }
    }
    cout << ans.size() << '\n';
    for (auto x : ans) cout << x << " ";
    return;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (!sieve[a[i] + a[j]]) {
        cout << 2 << " " << '\n';
        cout << a[i] << " " << a[j] << '\n';
        return;
      }
    }
  }
  cout << 1 << '\n';
  cout << a[1] << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Generate 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 = 1010;
const int K = int(2e6) + 10;
const int MOD = int(1e9) + 7;
const int INF = int(2e9) + 5;
const long long INF64 = 1e18;
int a[N];
int sieve[K];
void calc_sieve(int n) {
  sieve[1] = 1;
  for (int i = 2; i <= n; i++) {
    if (sieve[i]) continue;
    for (int u = 2 * i; u <= n; u += i) {
      sieve[u] = i;
    }
  }
}
void solve() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  calc_sieve(2 * a[n]);
  if (a[1] == 1 && a[2] == 1) {
    vector<int> ans;
    ans.push_back(1);
    for (int i = 2; i <= n; i++) {
      if (a[i] == 1) {
        ans.push_back(1);
        continue;
      }
      if (a[i] % 2 == 0 && sieve[1 + a[i]] == 0) {
        ans.push_back(a[i]);
        break;
      }
    }
    cout << ans.size() << '\n';
    for (auto x : ans) cout << x << " ";
    return;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (!sieve[a[i] + a[j]]) {
        cout << 2 << " " << '\n';
        cout << a[i] << " " << a[j] << '\n';
        return;
      }
    }
  }
  cout << 1 << '\n';
  cout << a[1] << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```