#include <bits/stdc++.h>
using namespace std;
const int N = 2 * (1e6) + 2;
bool is_prime[N + 77];
void sieve() {
  memset(is_prime, 1, sizeof is_prime);
  is_prime[0] = is_prime[1] = 0;
  for (int i = 4; i <= N; i += 2) is_prime[i] = 0;
  for (long long i = 3; i <= N; i += 2) {
    if (is_prime[i])
      for (long long j = i * i; j <= N; j += i) {
        is_prime[j] = 0;
      }
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  sieve();
  int n;
  cin >> n;
  vector<int> nums(n);
  int cnt = 0, ex = -1, one = -1;
  for (int i = 0; i < n; i++) {
    cin >> nums[i];
    if (nums[i] == 1) cnt += 1;
    if (nums[i] != 1 && is_prime[nums[i] + 1] && ex == -1) {
      cnt += 1;
      ex = nums[i];
    }
    if (is_prime[nums[i]]) one = nums[i];
  }
  int a = -1, b = -1;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (is_prime[nums[i] + nums[j]]) {
        a = nums[i];
        b = nums[j];
      }
    }
  }
  if (a != -1 && cnt < 2) {
    cout << "2\n";
    cout << a << " " << b << '\n';
  } else if (cnt == 1 && ex == -1) {
    cout << "1\n1\n";
  } else if (cnt >= 2) {
    cout << cnt << '\n';
    for (int i = ex == -1 ? 0 : 1; i < cnt; i++) cout << "1 ";
    if (ex != -1)
      cout << ex << '\n';
    else
      cout << '\n';
  } else if (one != -1) {
    cout << "1\n" << one << '\n';
  } else
    cout << "1\n" << nums[0] << '\n';
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 = 2 * (1e6) + 2;
bool is_prime[N + 77];
void sieve() {
  memset(is_prime, 1, sizeof is_prime);
  is_prime[0] = is_prime[1] = 0;
  for (int i = 4; i <= N; i += 2) is_prime[i] = 0;
  for (long long i = 3; i <= N; i += 2) {
    if (is_prime[i])
      for (long long j = i * i; j <= N; j += i) {
        is_prime[j] = 0;
      }
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  sieve();
  int n;
  cin >> n;
  vector<int> nums(n);
  int cnt = 0, ex = -1, one = -1;
  for (int i = 0; i < n; i++) {
    cin >> nums[i];
    if (nums[i] == 1) cnt += 1;
    if (nums[i] != 1 && is_prime[nums[i] + 1] && ex == -1) {
      cnt += 1;
      ex = nums[i];
    }
    if (is_prime[nums[i]]) one = nums[i];
  }
  int a = -1, b = -1;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (is_prime[nums[i] + nums[j]]) {
        a = nums[i];
        b = nums[j];
      }
    }
  }
  if (a != -1 && cnt < 2) {
    cout << "2\n";
    cout << a << " " << b << '\n';
  } else if (cnt == 1 && ex == -1) {
    cout << "1\n1\n";
  } else if (cnt >= 2) {
    cout << cnt << '\n';
    for (int i = ex == -1 ? 0 : 1; i < cnt; i++) cout << "1 ";
    if (ex != -1)
      cout << ex << '\n';
    else
      cout << '\n';
  } else if (one != -1) {
    cout << "1\n" << one << '\n';
  } else
    cout << "1\n" << nums[0] << '\n';
  return 0;
}

```