#include <bits/stdc++.h>
using namespace std;
bool prime_check(long long num) {
  if (num < 2) return 0;
  if (num % 2 == 0 && num != 2) return 0;
  for (int i = 3; i * i <= num; i++) {
    if (num % i == 0) return 0;
  }
  return 1;
}
void solve() {
  int n;
  cin >> n;
  int ctr = 0;
  int x;
  set<int> ev, od;
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    if (x == 1) {
      ctr++;
    }
    if (x & 1) {
      od.insert(x);
    } else {
      ev.insert(x);
    }
  }
  if (ctr) {
    int f = (ctr > 1);
    int x = 0;
    for (auto v : ev) {
      if (prime_check(1 + v)) {
        x = v;
        f = 1;
        break;
      }
    }
    if (ev.find(2) != ev.end()) {
      x = 2;
      f = 1;
    }
    if (f) {
      cout << ctr + (x != 0) << endl;
      for (int i = 0; i < ctr; i++) {
        cout << "1 ";
      }
      if (x) {
        cout << x;
      }
      cout << endl;
      return;
    }
  }
  for (auto u : od) {
    for (auto v : ev) {
      if (prime_check(u + v)) {
        cout << 2 << endl;
        cout << u << " " << v << endl;
        return;
      }
    }
  }
  cout << 1 << endl << x << endl;
}
int main(int argc, const char **argv) {
  if (argc >= 2) {
    freopen(argv[1], "r", stdin);
  }
  solve();
}

### Prompt

Generate a CPP solution to the following problem:

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;
bool prime_check(long long num) {
  if (num < 2) return 0;
  if (num % 2 == 0 && num != 2) return 0;
  for (int i = 3; i * i <= num; i++) {
    if (num % i == 0) return 0;
  }
  return 1;
}
void solve() {
  int n;
  cin >> n;
  int ctr = 0;
  int x;
  set<int> ev, od;
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    if (x == 1) {
      ctr++;
    }
    if (x & 1) {
      od.insert(x);
    } else {
      ev.insert(x);
    }
  }
  if (ctr) {
    int f = (ctr > 1);
    int x = 0;
    for (auto v : ev) {
      if (prime_check(1 + v)) {
        x = v;
        f = 1;
        break;
      }
    }
    if (ev.find(2) != ev.end()) {
      x = 2;
      f = 1;
    }
    if (f) {
      cout << ctr + (x != 0) << endl;
      for (int i = 0; i < ctr; i++) {
        cout << "1 ";
      }
      if (x) {
        cout << x;
      }
      cout << endl;
      return;
    }
  }
  for (auto u : od) {
    for (auto v : ev) {
      if (prime_check(u + v)) {
        cout << 2 << endl;
        cout << u << " " << v << endl;
        return;
      }
    }
  }
  cout << 1 << endl << x << endl;
}
int main(int argc, const char **argv) {
  if (argc >= 2) {
    freopen(argv[1], "r", stdin);
  }
  solve();
}

```