#include <bits/stdc++.h>
using namespace std;
bool prime[4000000];
int main() {
  fill(prime, prime + 4000000, true);
  prime[0] = false;
  prime[1] = false;
  for (int i = 2; i < 2000; i++) {
    if (!prime[i]) continue;
    for (int j = i * 2; j < 4000000; j += i) prime[j] = false;
  }
  int n;
  scanf("%d", &n);
  int ones = 0, notone = 1;
  int a[1000];
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    if (a[i] == 1)
      ones++;
    else if (prime[a[i] + 1])
      notone = a[i];
  }
  if (notone == 1) ones--;
  if (ones >= 1) {
    cout << ones + 1 << endl;
    for (int i = 0; i < ones; i++) cout << "1 ";
    cout << notone << endl;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (prime[a[i] + a[j]]) {
        cout << 2 << endl << a[i] << ' ' << a[j] << endl;
        return 0;
      }
    }
  }
  cout << 1 << endl << a[0] << endl;
}

### 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;
bool prime[4000000];
int main() {
  fill(prime, prime + 4000000, true);
  prime[0] = false;
  prime[1] = false;
  for (int i = 2; i < 2000; i++) {
    if (!prime[i]) continue;
    for (int j = i * 2; j < 4000000; j += i) prime[j] = false;
  }
  int n;
  scanf("%d", &n);
  int ones = 0, notone = 1;
  int a[1000];
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    if (a[i] == 1)
      ones++;
    else if (prime[a[i] + 1])
      notone = a[i];
  }
  if (notone == 1) ones--;
  if (ones >= 1) {
    cout << ones + 1 << endl;
    for (int i = 0; i < ones; i++) cout << "1 ";
    cout << notone << endl;
    return 0;
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (prime[a[i] + a[j]]) {
        cout << 2 << endl << a[i] << ' ' << a[j] << endl;
        return 0;
      }
    }
  }
  cout << 1 << endl << a[0] << endl;
}

```