#include <bits/stdc++.h>
using namespace std;
int n, a[2000], r[3000000];
void calc() {
  for (int i = 0; i < 3000000; i++) {
    r[i] = 0;
  }
  for (int i = 2; i < 1000001; i++) {
    int k = 2;
    while (i * k < 3000000) {
      r[i * k] = 1;
      k++;
    }
  }
}
pair<int, int> two() {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i != j && !r[a[i] + a[j]]) {
        return make_pair(a[i], a[j]);
      }
    }
  }
  return make_pair(-1, -1);
}
vector<int> one() {
  vector<int> ans;
  sort(a, a + n);
  int i = 0;
  while (i < n && a[i] == 1) {
    ans.push_back(1);
    i++;
  }
  int third = -1;
  for (int i = 2; i < n; i++) {
    if (!r[a[i] + 1] && ans.size() > 0 && a[i] != 1) {
      third = a[i];
    }
  }
  if (third > 0) {
    ans.push_back(third);
  }
  return ans;
}
int main() {
  calc();
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  pair<int, int> t = two();
  vector<int> with_one = one();
  if (with_one.size() > 2) {
    cout << with_one.size() << '\n';
    for (int i = 0; i < with_one.size(); i++) {
      cout << with_one[i] << ' ';
    }
    return 0;
  }
  if (t.first != -1) {
    cout << "2\n" << t.first << ' ' << t.second;
    return 0;
  }
  cout << 1 << '\n' << a[0];
  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;
int n, a[2000], r[3000000];
void calc() {
  for (int i = 0; i < 3000000; i++) {
    r[i] = 0;
  }
  for (int i = 2; i < 1000001; i++) {
    int k = 2;
    while (i * k < 3000000) {
      r[i * k] = 1;
      k++;
    }
  }
}
pair<int, int> two() {
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i != j && !r[a[i] + a[j]]) {
        return make_pair(a[i], a[j]);
      }
    }
  }
  return make_pair(-1, -1);
}
vector<int> one() {
  vector<int> ans;
  sort(a, a + n);
  int i = 0;
  while (i < n && a[i] == 1) {
    ans.push_back(1);
    i++;
  }
  int third = -1;
  for (int i = 2; i < n; i++) {
    if (!r[a[i] + 1] && ans.size() > 0 && a[i] != 1) {
      third = a[i];
    }
  }
  if (third > 0) {
    ans.push_back(third);
  }
  return ans;
}
int main() {
  calc();
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  pair<int, int> t = two();
  vector<int> with_one = one();
  if (with_one.size() > 2) {
    cout << with_one.size() << '\n';
    for (int i = 0; i < with_one.size(); i++) {
      cout << with_one[i] << ' ';
    }
    return 0;
  }
  if (t.first != -1) {
    cout << "2\n" << t.first << ' ' << t.second;
    return 0;
  }
  cout << 1 << '\n' << a[0];
  return 0;
}

```