#include <bits/stdc++.h>
using namespace std;
template <typename... Args>
void dbg(Args... args) {
  ((cout << args << " "), ...);
}
const int N = 3e6;
bitset<N> p;
int a[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  p[1] = p[0] = 1;
  for (int i = 4; i < N; i += 2) p[i] = 1;
  for (int i = 3; i * i < N; i += 2)
    if (!p[i])
      for (int j = i * i; j < N; j += i) p[j] = 1;
  int n;
  cin >> n;
  bool ok = 0;
  int cnt = 0;
  vector<int> ans1;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (a[i] == 1)
      ans1.emplace_back(1);
    else if (!p[a[i] + 1] and !ok) {
      ans1.emplace_back(a[i]);
      ok = 1;
    }
  }
  vector<int> ans2;
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j)
      if (!p[a[i] + a[j]]) {
        ans2 = {a[i], a[j]};
        break;
      }
  if (ans2.size() == 0) {
    ans2 = {a[1]};
  }
  if (ans1.size() >= ans2.size()) {
    dbg(ans1.size());
    for (int i : ans1) dbg(i);
  } else if (ans2.size() > ans1.size()) {
    dbg(ans2.size());
    for (int i : ans2) dbg(i);
  }
}

### Prompt

In Cpp, your task is to solve 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;
template <typename... Args>
void dbg(Args... args) {
  ((cout << args << " "), ...);
}
const int N = 3e6;
bitset<N> p;
int a[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  p[1] = p[0] = 1;
  for (int i = 4; i < N; i += 2) p[i] = 1;
  for (int i = 3; i * i < N; i += 2)
    if (!p[i])
      for (int j = i * i; j < N; j += i) p[j] = 1;
  int n;
  cin >> n;
  bool ok = 0;
  int cnt = 0;
  vector<int> ans1;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (a[i] == 1)
      ans1.emplace_back(1);
    else if (!p[a[i] + 1] and !ok) {
      ans1.emplace_back(a[i]);
      ok = 1;
    }
  }
  vector<int> ans2;
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j)
      if (!p[a[i] + a[j]]) {
        ans2 = {a[i], a[j]};
        break;
      }
  if (ans2.size() == 0) {
    ans2 = {a[1]};
  }
  if (ans1.size() >= ans2.size()) {
    dbg(ans1.size());
    for (int i : ans1) dbg(i);
  } else if (ans2.size() > ans1.size()) {
    dbg(ans2.size());
    for (int i : ans2) dbg(i);
  }
}

```