#include <bits/stdc++.h>
using namespace std;
long long n;
long long res;
long long unos;
long long v[1002];
vector<long long> resp;
bool comp[7010002];
void criba() {
  comp[0] = comp[1] = true;
  for (int i = 4; i < 7010002; i += 2) {
    comp[i] = true;
  }
  for (int i = 3; i < 7010002; i += 2) {
    if (!comp[i]) {
      for (int j = i + i; j < 7010002; j += i) {
        comp[j] = true;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  criba();
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    if (v[i] == 1) unos++;
  }
  for (int i = 0; i < n; i++) {
    if (1 > res) {
      res = 1;
      resp.push_back(v[i]);
    }
    if (!comp[v[i] + 1] && v[i] > 1 && unos) {
      if (unos + 1 > res) {
        res = 1 + unos;
        resp.clear();
        for (int j = 0; j < unos; j++) resp.push_back(1);
        resp.push_back(v[i]);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (v[i] == 1 || v[j] == 1) continue;
      if (!comp[v[i] + v[j] + 1] && unos && !comp[v[i] + v[j]]) {
        if (2 + unos > res) {
          res = 2 + unos;
          resp.clear();
          for (int it = 0; it < unos; it++) resp.push_back(1);
          resp.push_back(v[i]);
          resp.push_back(v[j]);
        }
      }
      if (!comp[v[i] + v[j]]) {
        if (2 > res) {
          res = 2;
          resp.clear();
          resp.push_back(v[i]);
          resp.push_back(v[j]);
        }
      }
    }
  }
  if (unos > res) {
    resp.clear();
    res = unos;
    for (int it = 0; it < unos; it++) resp.push_back(1);
  }
  cout << res << "\n";
  for (int i = 0; i < res; i++) {
    cout << resp[i] << " ";
  }
  cout << "\n";
}

### Prompt

Develop a solution in Cpp 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;
long long n;
long long res;
long long unos;
long long v[1002];
vector<long long> resp;
bool comp[7010002];
void criba() {
  comp[0] = comp[1] = true;
  for (int i = 4; i < 7010002; i += 2) {
    comp[i] = true;
  }
  for (int i = 3; i < 7010002; i += 2) {
    if (!comp[i]) {
      for (int j = i + i; j < 7010002; j += i) {
        comp[j] = true;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  criba();
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    if (v[i] == 1) unos++;
  }
  for (int i = 0; i < n; i++) {
    if (1 > res) {
      res = 1;
      resp.push_back(v[i]);
    }
    if (!comp[v[i] + 1] && v[i] > 1 && unos) {
      if (unos + 1 > res) {
        res = 1 + unos;
        resp.clear();
        for (int j = 0; j < unos; j++) resp.push_back(1);
        resp.push_back(v[i]);
      }
    }
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (v[i] == 1 || v[j] == 1) continue;
      if (!comp[v[i] + v[j] + 1] && unos && !comp[v[i] + v[j]]) {
        if (2 + unos > res) {
          res = 2 + unos;
          resp.clear();
          for (int it = 0; it < unos; it++) resp.push_back(1);
          resp.push_back(v[i]);
          resp.push_back(v[j]);
        }
      }
      if (!comp[v[i] + v[j]]) {
        if (2 > res) {
          res = 2;
          resp.clear();
          resp.push_back(v[i]);
          resp.push_back(v[j]);
        }
      }
    }
  }
  if (unos > res) {
    resp.clear();
    res = unos;
    for (int it = 0; it < unos; it++) resp.push_back(1);
  }
  cout << res << "\n";
  for (int i = 0; i < res; i++) {
    cout << resp[i] << " ";
  }
  cout << "\n";
}

```