#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  vector<bool> sieve(3000000, true);
  sieve[0] = false;
  sieve[1] = false;
  for (int i = 2; i <= 2000; i++) {
    if (!sieve[i]) continue;
    int sum = i + i;
    while (sum <= 2000000) {
      sieve[sum] = false;
      sum += i;
    }
  }
  int N;
  cin >> N;
  vector<int> nums(N);
  int numOnes = 0;
  bool hasGoodEven = false;
  int goodEven = 0;
  for (int i = 0; i < N; i++) {
    cin >> nums[i];
    if (nums[i] == 1)
      numOnes++;
    else if (sieve[nums[i] + 1]) {
      hasGoodEven = true;
      goodEven = nums[i];
    }
  }
  if (numOnes + hasGoodEven >= 2) {
    cout << numOnes + hasGoodEven << endl;
    for (int i = 0; i < numOnes; i++) {
      cout << "1 ";
    }
    if (hasGoodEven) cout << goodEven;
    cout << endl;
    return 0;
  }
  vector<int> res;
  for (int i = 0; i < N - 1; i++) {
    for (int j = i + 1; j < N; j++) {
      int curSum = nums[i] + nums[j];
      if (sieve[curSum]) {
        cout << "2" << endl;
        cout << nums[i] << " " << nums[j] << endl;
        return 0;
      }
    }
  }
  cout << "1" << endl;
  cout << nums[0] << endl;
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 main() {
  ios::sync_with_stdio(false);
  vector<bool> sieve(3000000, true);
  sieve[0] = false;
  sieve[1] = false;
  for (int i = 2; i <= 2000; i++) {
    if (!sieve[i]) continue;
    int sum = i + i;
    while (sum <= 2000000) {
      sieve[sum] = false;
      sum += i;
    }
  }
  int N;
  cin >> N;
  vector<int> nums(N);
  int numOnes = 0;
  bool hasGoodEven = false;
  int goodEven = 0;
  for (int i = 0; i < N; i++) {
    cin >> nums[i];
    if (nums[i] == 1)
      numOnes++;
    else if (sieve[nums[i] + 1]) {
      hasGoodEven = true;
      goodEven = nums[i];
    }
  }
  if (numOnes + hasGoodEven >= 2) {
    cout << numOnes + hasGoodEven << endl;
    for (int i = 0; i < numOnes; i++) {
      cout << "1 ";
    }
    if (hasGoodEven) cout << goodEven;
    cout << endl;
    return 0;
  }
  vector<int> res;
  for (int i = 0; i < N - 1; i++) {
    for (int j = i + 1; j < N; j++) {
      int curSum = nums[i] + nums[j];
      if (sieve[curSum]) {
        cout << "2" << endl;
        cout << nums[i] << " " << nums[j] << endl;
        return 0;
      }
    }
  }
  cout << "1" << endl;
  cout << nums[0] << endl;
  return 0;
}

```