#include <bits/stdc++.h>
using namespace std;
vector<long long> res;
long long* p = new long long[2000001];
void SNT() {
  for (long long i = 2; i * i < 2000001; i++) p[i] = 0;
  for (long long i = 2; i * i < 2000001; i++)
    if (!p[i])
      for (long long j = i * i; j < 2000001; j += i) p[j] = 1;
}
int main() {
  SNT();
  long long n;
  long long a[10001];
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  if (n > 1 && a[1] == 1) {
    long long i;
    for (i = 0; i < n && a[i] == 1; i++) res.push_back(1);
    for (i; i < n; i++) {
      if (!p[a[i] + 1]) break;
    }
    if (i < n) res.push_back(a[i]);
  } else {
    for (long long i = 0; i < n && res.empty(); i++) {
      for (long long j = i + 1; j < n && res.empty(); j++) {
        if (!p[a[i] + a[j]]) res.push_back(a[i]), res.push_back(a[j]);
      }
    }
    if (res.empty()) res.push_back(a[0]);
  }
  cout << (long long)res.size() << endl;
  for (long long i = 0; i < res.size(); i++) {
    cout << res.at(i) << " ";
  }
  return 0;
}

### 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;
vector<long long> res;
long long* p = new long long[2000001];
void SNT() {
  for (long long i = 2; i * i < 2000001; i++) p[i] = 0;
  for (long long i = 2; i * i < 2000001; i++)
    if (!p[i])
      for (long long j = i * i; j < 2000001; j += i) p[j] = 1;
}
int main() {
  SNT();
  long long n;
  long long a[10001];
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a, a + n);
  if (n > 1 && a[1] == 1) {
    long long i;
    for (i = 0; i < n && a[i] == 1; i++) res.push_back(1);
    for (i; i < n; i++) {
      if (!p[a[i] + 1]) break;
    }
    if (i < n) res.push_back(a[i]);
  } else {
    for (long long i = 0; i < n && res.empty(); i++) {
      for (long long j = i + 1; j < n && res.empty(); j++) {
        if (!p[a[i] + a[j]]) res.push_back(a[i]), res.push_back(a[j]);
      }
    }
    if (res.empty()) res.push_back(a[0]);
  }
  cout << (long long)res.size() << endl;
  for (long long i = 0; i < res.size(); i++) {
    cout << res.at(i) << " ";
  }
  return 0;
}

```