#include <bits/stdc++.h>
using namespace std;
int v[1000 + 10], n, nr1, n1;
bool ciur[2000000 + 10];
void sieve() {
  ciur[0] = ciur[1] = 1;
  for (int i = 2; i * i <= 2000000; i++)
    if (!ciur[i])
      for (int j = i * i; j <= 2000000; j += i) ciur[j] = 1;
}
int main() {
  sieve();
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
    if (v[i] == 1) nr1++;
  }
  if (nr1) {
    n1 = -1;
    for (int i = 1; i <= n; i++) {
      if (v[i] == 1) continue;
      if (!ciur[v[i] + 1]) {
        n1 = v[i];
        nr1++;
        break;
      }
    }
    if (nr1 >= 2) {
      cout << nr1 << '\n';
      for (int i = 1; i < nr1; i++) cout << 1 << ' ';
      if (n1 == -1)
        cout << 1 << '\n';
      else
        cout << n1 << '\n';
      return 0;
    }
  }
  int ok = 0;
  for (int i = 1; i <= n && !ok; i++) {
    if (v[i] == 1) continue;
    for (int j = i + 1; j <= n && !ok; j++)
      if (v[j] != 1 && !ciur[v[i] + v[j]]) {
        cout << 2 << '\n' << v[i] << ' ' << v[j] << '\n';
        return 0;
      }
  }
  cout << 1 << '\n' << v[1] << '\n';
  return 0;
}

### Prompt

Your task is to create 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 v[1000 + 10], n, nr1, n1;
bool ciur[2000000 + 10];
void sieve() {
  ciur[0] = ciur[1] = 1;
  for (int i = 2; i * i <= 2000000; i++)
    if (!ciur[i])
      for (int j = i * i; j <= 2000000; j += i) ciur[j] = 1;
}
int main() {
  sieve();
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> v[i];
    if (v[i] == 1) nr1++;
  }
  if (nr1) {
    n1 = -1;
    for (int i = 1; i <= n; i++) {
      if (v[i] == 1) continue;
      if (!ciur[v[i] + 1]) {
        n1 = v[i];
        nr1++;
        break;
      }
    }
    if (nr1 >= 2) {
      cout << nr1 << '\n';
      for (int i = 1; i < nr1; i++) cout << 1 << ' ';
      if (n1 == -1)
        cout << 1 << '\n';
      else
        cout << n1 << '\n';
      return 0;
    }
  }
  int ok = 0;
  for (int i = 1; i <= n && !ok; i++) {
    if (v[i] == 1) continue;
    for (int j = i + 1; j <= n && !ok; j++)
      if (v[j] != 1 && !ciur[v[i] + v[j]]) {
        cout << 2 << '\n' << v[i] << ' ' << v[j] << '\n';
        return 0;
      }
  }
  cout << 1 << '\n' << v[1] << '\n';
  return 0;
}

```