#include <bits/stdc++.h>
using namespace std;
int main() {
  const int LIM = 2000100;
  bool errat[LIM] = {};
  errat[1] = true;
  for (int i = 2; i < LIM; ++i) {
    if (!errat[i]) {
      for (int j = i + i; j < LIM; j += i) {
        errat[j] = true;
      }
    }
  }
  int n;
  cin >> n;
  vector<int> nums(n);
  for (int i = 0; i < n; ++i) {
    cin >> nums[i];
  }
  int p = -1;
  int p1 = -1, p2 = -1;
  for (int i = 0; i < nums.size(); ++i) {
    if (nums[i] > 1 && !errat[nums[i] + 1]) {
      p = nums[i];
    }
    for (int j = i + 1; j < nums.size(); ++j) {
      auto k = nums[i] + nums[j];
      if (!errat[k]) {
        p1 = nums[i];
        p2 = nums[j];
      }
    }
  }
  const auto one_cnt =
      count_if(nums.begin(), nums.end(), [](int k) { return k == 1; });
  if (one_cnt >= 2) {
    cout << one_cnt + (p != -1) << endl;
    for (int i = 0; i < one_cnt; ++i) {
      cout << "1 ";
    }
    if (p != -1) {
      cout << p << endl;
    }
  } else {
    if (p2 != -1) {
      cout << 2 << endl << p1 << ' ' << p2 << endl;
    } else {
      cout << 1 << endl << nums[0] << endl;
    }
  }
}

### 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() {
  const int LIM = 2000100;
  bool errat[LIM] = {};
  errat[1] = true;
  for (int i = 2; i < LIM; ++i) {
    if (!errat[i]) {
      for (int j = i + i; j < LIM; j += i) {
        errat[j] = true;
      }
    }
  }
  int n;
  cin >> n;
  vector<int> nums(n);
  for (int i = 0; i < n; ++i) {
    cin >> nums[i];
  }
  int p = -1;
  int p1 = -1, p2 = -1;
  for (int i = 0; i < nums.size(); ++i) {
    if (nums[i] > 1 && !errat[nums[i] + 1]) {
      p = nums[i];
    }
    for (int j = i + 1; j < nums.size(); ++j) {
      auto k = nums[i] + nums[j];
      if (!errat[k]) {
        p1 = nums[i];
        p2 = nums[j];
      }
    }
  }
  const auto one_cnt =
      count_if(nums.begin(), nums.end(), [](int k) { return k == 1; });
  if (one_cnt >= 2) {
    cout << one_cnt + (p != -1) << endl;
    for (int i = 0; i < one_cnt; ++i) {
      cout << "1 ";
    }
    if (p != -1) {
      cout << p << endl;
    }
  } else {
    if (p2 != -1) {
      cout << 2 << endl << p1 << ' ' << p2 << endl;
    } else {
      cout << 1 << endl << nums[0] << endl;
    }
  }
}

```