#include <bits/stdc++.h>
using namespace std;
int main() {
  vector<char> isprime(2000001, 1);
  int n;
  cin >> n;
  vector<int> arr(n);
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  isprime[0] = 0;
  isprime[1] = 0;
  for (int i = 2; i < 2000001; i++) {
    if (isprime[i]) {
      if (i * 1ll * i < 2000001) {
        for (int j = i * i; j < 2000001; j += i) {
          isprime[j] = 0;
        }
      }
    }
  }
  int a = 1999999, b = 1999999;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (isprime[arr[i] + arr[j]] &&
          (arr[i] == 1 || arr[j] == 1 || a == 1999999)) {
        a = arr[i];
        b = arr[j];
      }
    }
  }
  if (a == 1999999) {
    cout << 1 << '\n';
    cout << arr[0];
  } else {
    if (a == 1 || b == 1) {
      int res = 0;
      bool is = 1;
      for (int i = 0; i < n; i++) {
        if (arr[i] == 1) {
          res++;
        } else {
          if (is && isprime[arr[i] + 1]) {
            res++;
            is = 0;
          }
        }
      }
      cout << res << '\n';
      is = 1;
      for (int i = 0; i < n; i++) {
        if (arr[i] == 1) {
          cout << 1 << ' ';
        } else {
          if (is && isprime[arr[i] + 1]) {
            cout << arr[i] << ' ';
            is = 0;
          }
        }
      }
    } else {
      cout << 2 << '\n';
      cout << a << ' ' << b;
    }
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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 main() {
  vector<char> isprime(2000001, 1);
  int n;
  cin >> n;
  vector<int> arr(n);
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  isprime[0] = 0;
  isprime[1] = 0;
  for (int i = 2; i < 2000001; i++) {
    if (isprime[i]) {
      if (i * 1ll * i < 2000001) {
        for (int j = i * i; j < 2000001; j += i) {
          isprime[j] = 0;
        }
      }
    }
  }
  int a = 1999999, b = 1999999;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (isprime[arr[i] + arr[j]] &&
          (arr[i] == 1 || arr[j] == 1 || a == 1999999)) {
        a = arr[i];
        b = arr[j];
      }
    }
  }
  if (a == 1999999) {
    cout << 1 << '\n';
    cout << arr[0];
  } else {
    if (a == 1 || b == 1) {
      int res = 0;
      bool is = 1;
      for (int i = 0; i < n; i++) {
        if (arr[i] == 1) {
          res++;
        } else {
          if (is && isprime[arr[i] + 1]) {
            res++;
            is = 0;
          }
        }
      }
      cout << res << '\n';
      is = 1;
      for (int i = 0; i < n; i++) {
        if (arr[i] == 1) {
          cout << 1 << ' ';
        } else {
          if (is && isprime[arr[i] + 1]) {
            cout << arr[i] << ' ';
            is = 0;
          }
        }
      }
    } else {
      cout << 2 << '\n';
      cout << a << ' ' << b;
    }
  }
  return 0;
}

```