#include <bits/stdc++.h>
using namespace std;
vector<bool> primes(2000001, true);
bool isPrime(int num) { return primes[num]; }
void buildPrimes() {
  for (int i = 2; i <= 2000000; i++) {
    if (!primes[i]) {
      continue;
    }
    int j = 2 * i;
    while (j <= 2000000) {
      primes[j] = false;
      j += i;
    }
  }
}
int main() {
  cin.sync_with_stdio(0);
  int n;
  cin >> n;
  vector<int> nums(n);
  for (int i = 0; i < n; i++) {
    cin >> nums[i];
  }
  buildPrimes();
  sort(nums.begin(), nums.end());
  if (n == 1) {
    cout << 1 << endl;
    cout << nums[0] << endl;
    return 0;
  }
  int count = 0;
  for (int i = 0; i < n; i++) {
    if (nums[i] == 1) {
      count++;
      continue;
    }
    break;
  }
  if (count < 2) {
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        int sum = nums[i] + nums[j];
        if (isPrime(sum)) {
          cout << 2 << endl;
          cout << nums[i] << ' ' << nums[j] << endl;
          return 0;
        }
      }
    }
    cout << 1 << endl;
    cout << nums[0] << endl;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (nums[i] == 1) {
      continue;
    }
    int sum = nums[i] + 1;
    if (isPrime(sum)) {
      cout << count + 1 << endl;
      for (int i = 0; i < count; i++) {
        cout << "1 ";
      }
      cout << nums[i] << endl;
      return 0;
    }
  }
  cout << count << endl;
  for (int i = 0; i < count; i++) {
    cout << "1 ";
  }
  cout << endl;
  return 0;
  return 0;
}

### Prompt

Your challenge is to write 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;
vector<bool> primes(2000001, true);
bool isPrime(int num) { return primes[num]; }
void buildPrimes() {
  for (int i = 2; i <= 2000000; i++) {
    if (!primes[i]) {
      continue;
    }
    int j = 2 * i;
    while (j <= 2000000) {
      primes[j] = false;
      j += i;
    }
  }
}
int main() {
  cin.sync_with_stdio(0);
  int n;
  cin >> n;
  vector<int> nums(n);
  for (int i = 0; i < n; i++) {
    cin >> nums[i];
  }
  buildPrimes();
  sort(nums.begin(), nums.end());
  if (n == 1) {
    cout << 1 << endl;
    cout << nums[0] << endl;
    return 0;
  }
  int count = 0;
  for (int i = 0; i < n; i++) {
    if (nums[i] == 1) {
      count++;
      continue;
    }
    break;
  }
  if (count < 2) {
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        int sum = nums[i] + nums[j];
        if (isPrime(sum)) {
          cout << 2 << endl;
          cout << nums[i] << ' ' << nums[j] << endl;
          return 0;
        }
      }
    }
    cout << 1 << endl;
    cout << nums[0] << endl;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    if (nums[i] == 1) {
      continue;
    }
    int sum = nums[i] + 1;
    if (isPrime(sum)) {
      cout << count + 1 << endl;
      for (int i = 0; i < count; i++) {
        cout << "1 ";
      }
      cout << nums[i] << endl;
      return 0;
    }
  }
  cout << count << endl;
  for (int i = 0; i < count; i++) {
    cout << "1 ";
  }
  cout << endl;
  return 0;
  return 0;
}

```