#include <bits/stdc++.h>
using namespace std;
int A[1003];
int prime[2000006];
int main() {
  int N;
  cin >> N;
  prime[0] = prime[1] = 1;
  for (int i = 2; i <= 2000; i++)
    for (int j = i * i; j <= 2000000; j += i) prime[j] = 1;
  for (int i = 1; i <= N; i++) cin >> A[i];
  int ans = 0, one = -1, two = -1;
  for (int i = 1; i <= N; i++)
    if (!prime[A[i]]) ans = 1, one = A[i];
  for (int i = 1; i <= N; i++)
    for (int j = i + 1; j <= N; j++)
      if (!prime[A[i] + A[j]]) ans = 2, one = A[i], two = A[j];
  int ok = -1, xone = 0;
  for (int i = 1; i <= N; i++) {
    if (A[i] == 1)
      xone++;
    else if (!prime[A[i] + 1])
      ok = A[i];
  }
  if (xone == 0 || xone + (ok > -1) <= ans) {
    if (ans == 0) ans = 1, one = A[1];
    cout << ans << endl;
    if (ans == 1)
      cout << one << endl;
    else if (ans == 2)
      cout << one << " " << two << endl;
  } else {
    cout << xone + (ok > -1) << endl;
    for (int i = 0; i < xone; i++) cout << 1 << " ";
    if (ok > -1) cout << ok << endl;
  }
  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;
int A[1003];
int prime[2000006];
int main() {
  int N;
  cin >> N;
  prime[0] = prime[1] = 1;
  for (int i = 2; i <= 2000; i++)
    for (int j = i * i; j <= 2000000; j += i) prime[j] = 1;
  for (int i = 1; i <= N; i++) cin >> A[i];
  int ans = 0, one = -1, two = -1;
  for (int i = 1; i <= N; i++)
    if (!prime[A[i]]) ans = 1, one = A[i];
  for (int i = 1; i <= N; i++)
    for (int j = i + 1; j <= N; j++)
      if (!prime[A[i] + A[j]]) ans = 2, one = A[i], two = A[j];
  int ok = -1, xone = 0;
  for (int i = 1; i <= N; i++) {
    if (A[i] == 1)
      xone++;
    else if (!prime[A[i] + 1])
      ok = A[i];
  }
  if (xone == 0 || xone + (ok > -1) <= ans) {
    if (ans == 0) ans = 1, one = A[1];
    cout << ans << endl;
    if (ans == 1)
      cout << one << endl;
    else if (ans == 2)
      cout << one << " " << two << endl;
  } else {
    cout << xone + (ok > -1) << endl;
    for (int i = 0; i < xone; i++) cout << 1 << " ";
    if (ok > -1) cout << ok << endl;
  }
  return 0;
}

```