#include <bits/stdc++.h>
using namespace std;
vector<bool> erato(int n) {
  vector<bool> numbers(n + 1, true);
  numbers[0] = false;
  numbers[1] = false;
  for (int i = 2; i * i <= n; ++i) {
    if (numbers[i]) {
      for (int j = i * i; j <= n; j += i) {
        numbers[j] = false;
      }
    }
  }
  return numbers;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int a;
  cin >> a;
  vector<bool> prime = erato(2000010);
  vector<int> ans1;
  vector<int> ans2;
  vector<int> num(a);
  for (int i = 0; i < a; ++i) {
    cin >> num[i];
  }
  for (int i = 0; i < a - 1; ++i) {
    if (ans1.size() == 2) {
      break;
    }
    for (int j = i + 1; j < a; ++j) {
      if (prime[num[i] + num[j]]) {
        ans1.push_back(num[i]);
        ans1.push_back(num[j]);
        break;
      }
    }
  }
  bool flag = false;
  for (int i = 0; i < a; ++i) {
    if (num[i] == 1) {
      ans2.push_back(num[i]);
    } else if (prime[num[i] + 1] && !flag) {
      ans2.push_back(num[i]);
      flag = true;
    }
  }
  if (ans2.size() > 2) {
    cout << ans2.size() << "\n";
    for (int i = 0; i < ans2.size(); ++i) cout << ans2[i] << " ";
  } else if (ans1.size() == 2) {
    cout << "2"
         << "\n"
         << ans1[0] << " " << ans1[1];
  } else {
    cout << 1 << "\n" << num[0];
  }
  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;
vector<bool> erato(int n) {
  vector<bool> numbers(n + 1, true);
  numbers[0] = false;
  numbers[1] = false;
  for (int i = 2; i * i <= n; ++i) {
    if (numbers[i]) {
      for (int j = i * i; j <= n; j += i) {
        numbers[j] = false;
      }
    }
  }
  return numbers;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int a;
  cin >> a;
  vector<bool> prime = erato(2000010);
  vector<int> ans1;
  vector<int> ans2;
  vector<int> num(a);
  for (int i = 0; i < a; ++i) {
    cin >> num[i];
  }
  for (int i = 0; i < a - 1; ++i) {
    if (ans1.size() == 2) {
      break;
    }
    for (int j = i + 1; j < a; ++j) {
      if (prime[num[i] + num[j]]) {
        ans1.push_back(num[i]);
        ans1.push_back(num[j]);
        break;
      }
    }
  }
  bool flag = false;
  for (int i = 0; i < a; ++i) {
    if (num[i] == 1) {
      ans2.push_back(num[i]);
    } else if (prime[num[i] + 1] && !flag) {
      ans2.push_back(num[i]);
      flag = true;
    }
  }
  if (ans2.size() > 2) {
    cout << ans2.size() << "\n";
    for (int i = 0; i < ans2.size(); ++i) cout << ans2[i] << " ";
  } else if (ans1.size() == 2) {
    cout << "2"
         << "\n"
         << ans1[0] << " " << ans1[1];
  } else {
    cout << 1 << "\n" << num[0];
  }
  return 0;
}

```