#include <bits/stdc++.h>
using namespace std;
int n, a[1000009];
bool prime[2000009];
int main() {
  fill(prime + 2, prime + 2000005, true);
  for (int i = 2; i * i <= 2000000; i++) {
    if (!prime[i]) continue;
    for (int j = i * i; j <= 2000000; j += i) prime[j] = false;
  }
  cin >> n;
  vector<int> p;
  int c1 = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 1)
      c1++;
    else
      p.push_back(a[i]);
  }
  vector<int> ret = {a[0]};
  if (c1 >= 2) ret = vector<int>(c1, 1);
  for (int i : p) {
    if (prime[i + 1]) {
      vector<int> res(c1, 1);
      res.push_back(i);
      ret = res;
      break;
    }
  }
  bool flag1 = false;
  for (int i = 0; i < p.size() && !flag1; i++) {
    for (int j = i + 1; j < p.size() && !flag1; j++) {
      if (prime[p[i] + p[j]]) {
        if (ret.size() == 1) {
          ret = {p[i], p[j]};
          flag1 = true;
        }
      }
    }
  }
  bool flag2 = false;
  for (int i = 0; i < p.size() && !flag2; i++) {
    for (int j = i + 1; j < p.size() && !flag2; j++) {
      if (prime[p[i] + p[j]] && prime[1 + p[i]] && prime[1 + p[j]]) {
        if (ret.size() == 1) {
          vector<int> res(c1, 1);
          res.push_back(p[i]);
          res.push_back(p[j]);
          ret = res;
          flag2 = true;
        }
      }
    }
  }
  cout << ret.size() << endl;
  for (int i : ret) cout << i << ' ';
  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;
int n, a[1000009];
bool prime[2000009];
int main() {
  fill(prime + 2, prime + 2000005, true);
  for (int i = 2; i * i <= 2000000; i++) {
    if (!prime[i]) continue;
    for (int j = i * i; j <= 2000000; j += i) prime[j] = false;
  }
  cin >> n;
  vector<int> p;
  int c1 = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 1)
      c1++;
    else
      p.push_back(a[i]);
  }
  vector<int> ret = {a[0]};
  if (c1 >= 2) ret = vector<int>(c1, 1);
  for (int i : p) {
    if (prime[i + 1]) {
      vector<int> res(c1, 1);
      res.push_back(i);
      ret = res;
      break;
    }
  }
  bool flag1 = false;
  for (int i = 0; i < p.size() && !flag1; i++) {
    for (int j = i + 1; j < p.size() && !flag1; j++) {
      if (prime[p[i] + p[j]]) {
        if (ret.size() == 1) {
          ret = {p[i], p[j]};
          flag1 = true;
        }
      }
    }
  }
  bool flag2 = false;
  for (int i = 0; i < p.size() && !flag2; i++) {
    for (int j = i + 1; j < p.size() && !flag2; j++) {
      if (prime[p[i] + p[j]] && prime[1 + p[i]] && prime[1 + p[j]]) {
        if (ret.size() == 1) {
          vector<int> res(c1, 1);
          res.push_back(p[i]);
          res.push_back(p[j]);
          ret = res;
          flag2 = true;
        }
      }
    }
  }
  cout << ret.size() << endl;
  for (int i : ret) cout << i << ' ';
  return 0;
}

```