#include <bits/stdc++.h>
using namespace std;
const int N = 1005, M = 2e6 + 5;
int sieve[M], arr[N];
void pre() {
  int h = 2e6 + 2;
  int f = sqrt(h);
  for (int i = 3; i <= f; i++) {
    if (sieve[i]) continue;
    for (int j = i * 3; j <= h; j += i * 2) {
      sieve[j] = 1;
    }
  }
}
bool isPrime(int x) {
  if (x == 2)
    return true;
  else if (x % 2 == 0)
    return false;
  else
    return !sieve[x];
}
int main() {
  pre();
  int n;
  scanf("%d", &n);
  int ones = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", arr + i);
    if (arr[i] == 1) ones++;
  }
  if (ones > 1) {
    for (int i = 1; i <= n; i++) {
      if (arr[i] != 1 && isPrime(arr[i] + 1)) {
        cout << ones + 1 << endl;
        for (int j = 1; j <= ones; j++) {
          cout << 1 << " ";
        }
        cout << arr[i] << endl;
        return 0;
      }
    }
    cout << ones << endl;
    for (int j = 1; j <= ones; j++) {
      cout << 1 << " ";
    }
    cout << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (isPrime(arr[i] + arr[j])) {
        cout << 2 << endl << arr[i] << " " << arr[j] << endl;
        return 0;
      }
    }
  }
  cout << 1 << endl << arr[1] << endl;
  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 = 1005, M = 2e6 + 5;
int sieve[M], arr[N];
void pre() {
  int h = 2e6 + 2;
  int f = sqrt(h);
  for (int i = 3; i <= f; i++) {
    if (sieve[i]) continue;
    for (int j = i * 3; j <= h; j += i * 2) {
      sieve[j] = 1;
    }
  }
}
bool isPrime(int x) {
  if (x == 2)
    return true;
  else if (x % 2 == 0)
    return false;
  else
    return !sieve[x];
}
int main() {
  pre();
  int n;
  scanf("%d", &n);
  int ones = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", arr + i);
    if (arr[i] == 1) ones++;
  }
  if (ones > 1) {
    for (int i = 1; i <= n; i++) {
      if (arr[i] != 1 && isPrime(arr[i] + 1)) {
        cout << ones + 1 << endl;
        for (int j = 1; j <= ones; j++) {
          cout << 1 << " ";
        }
        cout << arr[i] << endl;
        return 0;
      }
    }
    cout << ones << endl;
    for (int j = 1; j <= ones; j++) {
      cout << 1 << " ";
    }
    cout << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (isPrime(arr[i] + arr[j])) {
        cout << 2 << endl << arr[i] << " " << arr[j] << endl;
        return 0;
      }
    }
  }
  cout << 1 << endl << arr[1] << endl;
  return 0;
}

```