#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
int mpow(int base, int exp);
const int N = 1e6 + 3;
bool prime[2 * N];
void pre(int n = 2 * N) {
  memset(prime, true, sizeof(prime));
  prime[1] = false;
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * 2; i <= n; i += p) prime[i] = false;
    }
  }
}
int main() {
  int i, j, n;
  pre();
  cin >> n;
  int a[n], one, two;
  one = two = 0;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 1) one++;
    if (a[i] == 2) two++;
  }
  int A = -1, B, C = -1, D = -1;
  for (i = 0; i < n; i++) {
    if (a[i] != 1 && prime[1 + a[i]]) C = a[i];
    if (prime[a[i]]) D = a[i];
    for (j = i + 1; j < n; j++) {
      if (prime[a[i] + a[j]]) {
        A = a[i];
        B = a[j];
      }
    }
  }
  if (C == -1) {
    if (one >= 2) {
      cout << one << endl;
      for (i = 0; i < one; i++) cout << "1 ";
      return 0;
    }
  } else {
    if (one + 1 >= 2) {
      cout << one + 1 << endl;
      for (i = 0; i < one; i++) cout << "1 ";
      cout << C << endl;
      return 0;
    }
  }
  if (A != -1) {
    if (one >= 2) {
      cout << one << endl;
      for (i = 0; i < one; i++) cout << "1 ";
      return 0;
    }
    cout << 2 << endl;
    cout << A << " " << B << " " << endl;
    return 0;
  }
  if (D != -1) {
    if (one >= 2) {
      cout << one << endl;
      for (i = 0; i < one; i++) cout << "1 ";
      return 0;
    }
    cout << 1 << endl;
    cout << D << " " << endl;
    return 0;
  }
  cout << 1 << endl;
  cout << a[0] << endl;
  return 0;
}
int mpow(int base, int exp) {
  base %= mod;
  int result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((long long)result * base) % mod;
    base = ((long long)base * base) % mod;
    exp >>= 1;
  }
  return result;
}

### Prompt

Please provide a cpp coded solution 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;
const int mod = 1000000007;
int mpow(int base, int exp);
const int N = 1e6 + 3;
bool prime[2 * N];
void pre(int n = 2 * N) {
  memset(prime, true, sizeof(prime));
  prime[1] = false;
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * 2; i <= n; i += p) prime[i] = false;
    }
  }
}
int main() {
  int i, j, n;
  pre();
  cin >> n;
  int a[n], one, two;
  one = two = 0;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 1) one++;
    if (a[i] == 2) two++;
  }
  int A = -1, B, C = -1, D = -1;
  for (i = 0; i < n; i++) {
    if (a[i] != 1 && prime[1 + a[i]]) C = a[i];
    if (prime[a[i]]) D = a[i];
    for (j = i + 1; j < n; j++) {
      if (prime[a[i] + a[j]]) {
        A = a[i];
        B = a[j];
      }
    }
  }
  if (C == -1) {
    if (one >= 2) {
      cout << one << endl;
      for (i = 0; i < one; i++) cout << "1 ";
      return 0;
    }
  } else {
    if (one + 1 >= 2) {
      cout << one + 1 << endl;
      for (i = 0; i < one; i++) cout << "1 ";
      cout << C << endl;
      return 0;
    }
  }
  if (A != -1) {
    if (one >= 2) {
      cout << one << endl;
      for (i = 0; i < one; i++) cout << "1 ";
      return 0;
    }
    cout << 2 << endl;
    cout << A << " " << B << " " << endl;
    return 0;
  }
  if (D != -1) {
    if (one >= 2) {
      cout << one << endl;
      for (i = 0; i < one; i++) cout << "1 ";
      return 0;
    }
    cout << 1 << endl;
    cout << D << " " << endl;
    return 0;
  }
  cout << 1 << endl;
  cout << a[0] << endl;
  return 0;
}
int mpow(int base, int exp) {
  base %= mod;
  int result = 1;
  while (exp > 0) {
    if (exp & 1) result = ((long long)result * base) % mod;
    base = ((long long)base * base) % mod;
    exp >>= 1;
  }
  return result;
}

```