#include <bits/stdc++.h>
using namespace std;
long long int mod = 1000000007;
bool p[2000005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int test = 1;
  while (test--) {
    int n;
    cin >> n;
    int a[n];
    int c = 0;
    for (int i = (0); i < (n); i++) {
      cin >> a[i];
      if (a[i] == 1) c++;
    }
    for (int i = (2); i < (2000005); i++) {
      int t = 2;
      if (p[i] == 0) {
        while ((t * i) < 2000005) {
          p[i * t] = 1;
          t++;
        }
      }
    }
    for (int i = (0); i < (n); i++) {
      if (a[i] != 1) {
        if (c > 0) {
          if (p[a[i] + 1] == 0) {
            cout << c + 1 << '\n';
            for (int j = (0); j < (c); j++) cout << 1 << ' ';
            cout << a[i];
            return 0;
          }
        }
        if (c < 2) {
          for (int j = (0); j < (n); j++) {
            if (i != j && a[j] != 1 && p[a[j] + a[i]] == 0) {
              cout << 2 << '\n' << a[i] << ' ' << a[j];
              return 0;
            }
          }
        }
      }
    }
    if (c > 0) {
      cout << c << '\n';
      for (int i = (0); i < (c); i++) cout << 1 << ' ';
    } else {
      cout << 1 << '\n';
      cout << a[0];
    }
  }
  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;
long long int mod = 1000000007;
bool p[2000005];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int test = 1;
  while (test--) {
    int n;
    cin >> n;
    int a[n];
    int c = 0;
    for (int i = (0); i < (n); i++) {
      cin >> a[i];
      if (a[i] == 1) c++;
    }
    for (int i = (2); i < (2000005); i++) {
      int t = 2;
      if (p[i] == 0) {
        while ((t * i) < 2000005) {
          p[i * t] = 1;
          t++;
        }
      }
    }
    for (int i = (0); i < (n); i++) {
      if (a[i] != 1) {
        if (c > 0) {
          if (p[a[i] + 1] == 0) {
            cout << c + 1 << '\n';
            for (int j = (0); j < (c); j++) cout << 1 << ' ';
            cout << a[i];
            return 0;
          }
        }
        if (c < 2) {
          for (int j = (0); j < (n); j++) {
            if (i != j && a[j] != 1 && p[a[j] + a[i]] == 0) {
              cout << 2 << '\n' << a[i] << ' ' << a[j];
              return 0;
            }
          }
        }
      }
    }
    if (c > 0) {
      cout << c << '\n';
      for (int i = (0); i < (c); i++) cout << 1 << ' ';
    } else {
      cout << 1 << '\n';
      cout << a[0];
    }
  }
  return 0;
}

```