#include <bits/stdc++.h>
using namespace std;
long long prime[2000006];
int main() {
  long long i, j;
  prime[1] = 1;
  for (i = 2; i < 2000006; i++) {
    for (j = 2 * i; j < 2000006; j += i) prime[j] = 1;
  }
  long long n;
  cin >> n;
  long long a[n], co = 0;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 1) co++;
  }
  if (co > 1) {
    for (i = 0; i < n; i++) {
      if (prime[a[i] + 1] == 0 && a[i] > 1) {
        cout << co + 1 << endl << a[i] << " ";
        for (j = 0; j < co; j++) cout << 1 << " ";
        return 0;
      }
    }
    cout << co << endl;
    for (i = 0; i < co; i++) cout << 1 << " ";
    return 0;
  }
  for (i = 0; i < n; i++) {
    for (j = i + 1; j < n; j++) {
      if (prime[a[i] + a[j]] == 0) {
        cout << 2 << endl;
        cout << a[i] << " " << a[j];
        return 0;
      }
    }
  }
  cout << 1 << endl << a[0];
  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;
long long prime[2000006];
int main() {
  long long i, j;
  prime[1] = 1;
  for (i = 2; i < 2000006; i++) {
    for (j = 2 * i; j < 2000006; j += i) prime[j] = 1;
  }
  long long n;
  cin >> n;
  long long a[n], co = 0;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 1) co++;
  }
  if (co > 1) {
    for (i = 0; i < n; i++) {
      if (prime[a[i] + 1] == 0 && a[i] > 1) {
        cout << co + 1 << endl << a[i] << " ";
        for (j = 0; j < co; j++) cout << 1 << " ";
        return 0;
      }
    }
    cout << co << endl;
    for (i = 0; i < co; i++) cout << 1 << " ";
    return 0;
  }
  for (i = 0; i < n; i++) {
    for (j = i + 1; j < n; j++) {
      if (prime[a[i] + a[j]] == 0) {
        cout << 2 << endl;
        cout << a[i] << " " << a[j];
        return 0;
      }
    }
  }
  cout << 1 << endl << a[0];
  return 0;
}

```