#include <bits/stdc++.h>
using namespace std;
const int MX = 2 * 1e5 + 10;
const double PI = acos(-1.0), EPS = 1e-9;
int N;
int arr[2000];
map<int, int> mpp;
vector<pair<int, int> > best;
int p[2000000 + 2];
bool prime(int n) {
  if (n == 2) return 1;
  if (n % 2 == 0 || n < 2) return 0;
  for (int i = 3; i * i <= n; i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
int calc() {
  int ret = 0;
  for (int i = 0; i < best.size(); i++) ret += best[i].second;
  return ret;
}
void print() {
  int mx = calc();
  cout << mx << "\n";
  for (int i = 0; i < best.size(); i++)
    for (int j = 0; j < best[i].second; j++) cout << best[i].first << " ";
}
int main() {
  mpp.clear();
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d", &arr[i]), mpp[arr[i]]++;
  if (mpp.find(1) != mpp.end()) {
    best.push_back(make_pair(1, mpp[1]));
    for (int i = 0; i < N; i++) {
      if (arr[i] == 1) continue;
      if (prime(1 + arr[i])) {
        best.push_back(make_pair(arr[i], 1));
        break;
      }
    }
  }
  int maxbest = calc();
  if (maxbest > 2) {
    print();
    return 0;
  }
  best.clear();
  bool ok = 0;
  for (int i = 0; i < N; i++) {
    for (int j = i + 1; j < N; j++) {
      if (i == j) continue;
      if (prime(arr[i] + arr[j])) {
        best.push_back(make_pair(arr[i], 1));
        best.push_back(make_pair(arr[j], 1));
        ok = 1;
        break;
      }
    }
    if (ok) break;
  }
  if (best.size() == 0) best.push_back(make_pair(arr[0], 1));
  print();
  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;
const int MX = 2 * 1e5 + 10;
const double PI = acos(-1.0), EPS = 1e-9;
int N;
int arr[2000];
map<int, int> mpp;
vector<pair<int, int> > best;
int p[2000000 + 2];
bool prime(int n) {
  if (n == 2) return 1;
  if (n % 2 == 0 || n < 2) return 0;
  for (int i = 3; i * i <= n; i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
int calc() {
  int ret = 0;
  for (int i = 0; i < best.size(); i++) ret += best[i].second;
  return ret;
}
void print() {
  int mx = calc();
  cout << mx << "\n";
  for (int i = 0; i < best.size(); i++)
    for (int j = 0; j < best[i].second; j++) cout << best[i].first << " ";
}
int main() {
  mpp.clear();
  scanf("%d", &N);
  for (int i = 0; i < N; i++) scanf("%d", &arr[i]), mpp[arr[i]]++;
  if (mpp.find(1) != mpp.end()) {
    best.push_back(make_pair(1, mpp[1]));
    for (int i = 0; i < N; i++) {
      if (arr[i] == 1) continue;
      if (prime(1 + arr[i])) {
        best.push_back(make_pair(arr[i], 1));
        break;
      }
    }
  }
  int maxbest = calc();
  if (maxbest > 2) {
    print();
    return 0;
  }
  best.clear();
  bool ok = 0;
  for (int i = 0; i < N; i++) {
    for (int j = i + 1; j < N; j++) {
      if (i == j) continue;
      if (prime(arr[i] + arr[j])) {
        best.push_back(make_pair(arr[i], 1));
        best.push_back(make_pair(arr[j], 1));
        ok = 1;
        break;
      }
    }
    if (ok) break;
  }
  if (best.size() == 0) best.push_back(make_pair(arr[0], 1));
  print();
  return 0;
}

```