#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 5;
int a[N];
int h[N];
int p[N];
void pre() {
  p[1] = 1;
  for (int i = 2; i < N; i++) {
    if (p[i] == 0) {
      for (int j = i + i; j < N; j += i) p[j] = 1;
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  pre();
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    h[a[i]]++;
  }
  int xx = -1, yy = -1;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (!p[a[i] + a[j]]) {
        xx = a[i], yy = a[j];
      }
    }
  }
  int px = -1;
  for (int i = 0; i < n; i++) {
    if (a[i] != 1 && !p[a[i] + 1]) {
      px = a[i];
    }
  }
  vector<int> yo;
  if (xx != -1)
    yo = {xx, yy};
  else
    yo = {a[0]};
  vector<int> vo;
  for (int j = 0; j < h[1]; j++) {
    vo.push_back(1);
  }
  if (px != -1) vo.push_back(px);
  if ((int)yo.size() < (int)vo.size()) swap(yo, vo);
  cout << (int)yo.size() << "\n";
  for (auto i : yo) cout << i << " ";
  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 N = 2e6 + 5;
int a[N];
int h[N];
int p[N];
void pre() {
  p[1] = 1;
  for (int i = 2; i < N; i++) {
    if (p[i] == 0) {
      for (int j = i + i; j < N; j += i) p[j] = 1;
    }
  }
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  pre();
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    h[a[i]]++;
  }
  int xx = -1, yy = -1;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (!p[a[i] + a[j]]) {
        xx = a[i], yy = a[j];
      }
    }
  }
  int px = -1;
  for (int i = 0; i < n; i++) {
    if (a[i] != 1 && !p[a[i] + 1]) {
      px = a[i];
    }
  }
  vector<int> yo;
  if (xx != -1)
    yo = {xx, yy};
  else
    yo = {a[0]};
  vector<int> vo;
  for (int j = 0; j < h[1]; j++) {
    vo.push_back(1);
  }
  if (px != -1) vo.push_back(px);
  if ((int)yo.size() < (int)vo.size()) swap(yo, vo);
  cout << (int)yo.size() << "\n";
  for (auto i : yo) cout << i << " ";
  return 0;
}

```