#include <bits/stdc++.h>
using namespace std;
const int MAXN = 10000;
const int MAXK = 2000010;
bool isPrime[MAXK];
int A[MAXN];
void eratosfen() {
  fill(isPrime, isPrime + MAXK, true);
  for (int i = 2; i < MAXK; i++) {
    if (!isPrime[i]) continue;
    for (int j = i + i; j < MAXK; j += i) isPrime[j] = false;
  }
  isPrime[1] = 0;
  isPrime[0] = 0;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> A[i];
  }
  eratosfen();
  sort(A, A + n);
  int ones = 0;
  int evens = 0;
  for (int i = 0; i < n; ++i) {
    if (A[i] == 1) ones++;
    if (isPrime[A[i] + 1] && A[i] != 1) evens = A[i];
  }
  if (ones + (evens != 0) > 1) {
    cout << ones + (evens != 0) << endl;
    for (int i = 0; i < ones; ++i) cout << 1 << " ";
    if (evens) cout << evens;
    cout << endl;
    return 0;
  }
  for (int i = 0; i < n; ++i)
    for (int j = i + 1; j < n; ++j) {
      if (isPrime[A[i] + A[j]]) {
        cout << 2 << endl;
        cout << A[i] << " " << A[j] << endl;
        return 0;
      }
    }
  cout << 1 << endl;
  cout << A[0] << endl;
  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;
const int MAXN = 10000;
const int MAXK = 2000010;
bool isPrime[MAXK];
int A[MAXN];
void eratosfen() {
  fill(isPrime, isPrime + MAXK, true);
  for (int i = 2; i < MAXK; i++) {
    if (!isPrime[i]) continue;
    for (int j = i + i; j < MAXK; j += i) isPrime[j] = false;
  }
  isPrime[1] = 0;
  isPrime[0] = 0;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> A[i];
  }
  eratosfen();
  sort(A, A + n);
  int ones = 0;
  int evens = 0;
  for (int i = 0; i < n; ++i) {
    if (A[i] == 1) ones++;
    if (isPrime[A[i] + 1] && A[i] != 1) evens = A[i];
  }
  if (ones + (evens != 0) > 1) {
    cout << ones + (evens != 0) << endl;
    for (int i = 0; i < ones; ++i) cout << 1 << " ";
    if (evens) cout << evens;
    cout << endl;
    return 0;
  }
  for (int i = 0; i < n; ++i)
    for (int j = i + 1; j < n; ++j) {
      if (isPrime[A[i] + A[j]]) {
        cout << 2 << endl;
        cout << A[i] << " " << A[j] << endl;
        return 0;
      }
    }
  cout << 1 << endl;
  cout << A[0] << endl;
  return 0;
}

```