#include <bits/stdc++.h>
using namespace std;
int const N = 2e6 + 4;
vector<bool> is_prime(N + 5, true);
void seive() {
  is_prime[1] = is_prime[0] = false;
  for (long long i = 2; i <= N; i++) {
    if (is_prime[i]) {
      for (long long j = i * i; j <= N; j += i) is_prime[j] = false;
    }
  }
}
int frq[N];
int main() {
  seive();
  int n;
  cin >> n;
  int a[n];
  vector<int> res1;
  pair<int, int> res2 = {-1, -1};
  int res3 = -1;
  int mx = 0, num = -1;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    if (a[i] == 1) res1.push_back(1);
    if (is_prime[a[i]]) res3 = a[i];
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      if (is_prime[a[i] + a[j]]) res2 = {a[i], a[j]};
    }
    if (a[i] != 1 && is_prime[a[i] + 1]) {
      num = a[i];
    }
  }
  if (!res1.empty() && num != -1) res1.push_back(num);
  if (res1.empty() && res2.first == -1 && res3 == -1) {
    cout << 1 << endl;
    cout << a[0] << endl;
  } else if (res2.first == -1 || res1.size() > 2) {
    if (res1.empty()) {
      cout << 1 << endl << res3 << endl;
    } else {
      cout << res1.size() << endl;
      for (int ele : res1) cout << ele << " ";
    }
  } else {
    cout << 2 << endl;
    cout << res2.first << " " << res2.second << endl;
  }
}

### 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;
int const N = 2e6 + 4;
vector<bool> is_prime(N + 5, true);
void seive() {
  is_prime[1] = is_prime[0] = false;
  for (long long i = 2; i <= N; i++) {
    if (is_prime[i]) {
      for (long long j = i * i; j <= N; j += i) is_prime[j] = false;
    }
  }
}
int frq[N];
int main() {
  seive();
  int n;
  cin >> n;
  int a[n];
  vector<int> res1;
  pair<int, int> res2 = {-1, -1};
  int res3 = -1;
  int mx = 0, num = -1;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) {
    if (a[i] == 1) res1.push_back(1);
    if (is_prime[a[i]]) res3 = a[i];
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      if (is_prime[a[i] + a[j]]) res2 = {a[i], a[j]};
    }
    if (a[i] != 1 && is_prime[a[i] + 1]) {
      num = a[i];
    }
  }
  if (!res1.empty() && num != -1) res1.push_back(num);
  if (res1.empty() && res2.first == -1 && res3 == -1) {
    cout << 1 << endl;
    cout << a[0] << endl;
  } else if (res2.first == -1 || res1.size() > 2) {
    if (res1.empty()) {
      cout << 1 << endl << res3 << endl;
    } else {
      cout << res1.size() << endl;
      for (int ele : res1) cout << ele << " ";
    }
  } else {
    cout << 2 << endl;
    cout << res2.first << " " << res2.second << endl;
  }
}

```