#include <bits/stdc++.h>
using namespace std;
int n, i, a[11111], k, j, r, p[3333];
bool Prost(int q) {
  for (int j = 0; j < r; j++) {
    if (!(q % p[j]) && q != p[j]) return 0;
    if (p[j] * p[j] > q) break;
  }
  return 1;
}
int main() {
  for (i = 2; i <= sqrt(2000000); i++)
    if (Prost(i)) {
      p[r] = i;
      r++;
    }
  cin >> n;
  for (i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  if (a[0] == 1 && a[1] == 1) {
    i = 0;
    while (a[i] == 1) {
      i++;
      k++;
    }
    for (; i < n; i++) {
      if (Prost(a[i] + 1)) {
        cout << k + 1 << endl;
        cout << a[i] << " ";
        for (i = 0; i < k; i++) cout << 1 << " ";
        return 0;
      }
    }
    cout << k << endl;
    for (i = 0; i < k; i++) cout << 1 << " ";
    return 0;
  } else {
    for (i = 0; i < n; i++)
      for (j = i + 1; j < n; j++)
        if (Prost(a[i] + a[j])) {
          cout << 2 << endl;
          cout << a[i] << " " << a[j];
          return 0;
        }
    cout << 1 << endl << a[0];
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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;
int n, i, a[11111], k, j, r, p[3333];
bool Prost(int q) {
  for (int j = 0; j < r; j++) {
    if (!(q % p[j]) && q != p[j]) return 0;
    if (p[j] * p[j] > q) break;
  }
  return 1;
}
int main() {
  for (i = 2; i <= sqrt(2000000); i++)
    if (Prost(i)) {
      p[r] = i;
      r++;
    }
  cin >> n;
  for (i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  if (a[0] == 1 && a[1] == 1) {
    i = 0;
    while (a[i] == 1) {
      i++;
      k++;
    }
    for (; i < n; i++) {
      if (Prost(a[i] + 1)) {
        cout << k + 1 << endl;
        cout << a[i] << " ";
        for (i = 0; i < k; i++) cout << 1 << " ";
        return 0;
      }
    }
    cout << k << endl;
    for (i = 0; i < k; i++) cout << 1 << " ";
    return 0;
  } else {
    for (i = 0; i < n; i++)
      for (j = i + 1; j < n; j++)
        if (Prost(a[i] + a[j])) {
          cout << 2 << endl;
          cout << a[i] << " " << a[j];
          return 0;
        }
    cout << 1 << endl << a[0];
  }
  return 0;
}

```