#include <bits/stdc++.h>
const long long INF = 4e18L + 1;
const long long IINF = 2e9 + 1;
const long long limit = 1048576;
using namespace std;
long long n;
vector<long long> a;
bool added = false;
vector<long long> cand1, cand2;
const long long N = 2e6;
vector<bool> is_prime;
void Erasto() {
  is_prime.resize(N + 1, true);
  for (long long i = (long long)2; i < (long long)N + 1; i++) {
    if (is_prime[i]) {
      for (int j = 2 * i; j < N + 1; j += i) {
        is_prime[j] = false;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  Erasto();
  cin >> n;
  a.resize(n);
  for (auto& i : a) {
    cin >> i;
    if (i == 1) {
      cand1.push_back(i);
    } else if (!added and is_prime[i + 1]) {
      cand1.push_back(i);
      added = true;
    }
  }
  for (long long i = (long long)0; i < (long long)n; i++) {
    for (long long j = (long long)0; j < (long long)n; j++) {
      if (i != j and cand2.size() == 0 and is_prime[a[i] + a[j]]) {
        cand2 = {a[i], a[j]};
      }
    }
  }
  if (!cand2.size()) {
    cand2 = {a[0]};
  }
  if (cand1.size() > cand2.size()) {
    cout << cand1.size() << "\n";
    for (auto& i : cand1) {
      cout << i << " ";
    }
  } else {
    cout << cand2.size() << "\n";
    for (auto& i : cand2) {
      cout << i << " ";
    }
  }
  cout << "\n";
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
const long long INF = 4e18L + 1;
const long long IINF = 2e9 + 1;
const long long limit = 1048576;
using namespace std;
long long n;
vector<long long> a;
bool added = false;
vector<long long> cand1, cand2;
const long long N = 2e6;
vector<bool> is_prime;
void Erasto() {
  is_prime.resize(N + 1, true);
  for (long long i = (long long)2; i < (long long)N + 1; i++) {
    if (is_prime[i]) {
      for (int j = 2 * i; j < N + 1; j += i) {
        is_prime[j] = false;
      }
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  Erasto();
  cin >> n;
  a.resize(n);
  for (auto& i : a) {
    cin >> i;
    if (i == 1) {
      cand1.push_back(i);
    } else if (!added and is_prime[i + 1]) {
      cand1.push_back(i);
      added = true;
    }
  }
  for (long long i = (long long)0; i < (long long)n; i++) {
    for (long long j = (long long)0; j < (long long)n; j++) {
      if (i != j and cand2.size() == 0 and is_prime[a[i] + a[j]]) {
        cand2 = {a[i], a[j]};
      }
    }
  }
  if (!cand2.size()) {
    cand2 = {a[0]};
  }
  if (cand1.size() > cand2.size()) {
    cout << cand1.size() << "\n";
    for (auto& i : cand1) {
      cout << i << " ";
    }
  } else {
    cout << cand2.size() << "\n";
    for (auto& i : cand2) {
      cout << i << " ";
    }
  }
  cout << "\n";
  return 0;
}

```