#include <bits/stdc++.h>
using namespace std;
bool prime[3000000];
void precalc() {
  for (int i = 1; i < 3000000; i++) prime[i] = true;
  for (int i = 2; i < 3000000; i += 2) prime[i] = false;
  int p = sqrt(3000000);
  prime[2] = true;
  for (int i = 3; i <= p; i += 2) {
    for (int j = i * i; j < 3000000; j += 2 * i) {
      prime[j] = false;
    }
  }
}
int main() {
  int i, j, k, l, m, n, rem[1001];
  precalc();
  while (cin >> n) {
    int ara[n];
    m = 0;
    for (i = 0; i < n; i++) {
      cin >> ara[i];
      if (ara[i] == 1) m++;
    }
    if (m) {
      k = 0;
      for (i = 0; i < n; i++) {
        if (ara[i] != 1 && prime[ara[i] + 1]) {
          k = ara[i];
        }
      }
      if (k) {
        cout << m + 1 << endl;
        for (i = 0; i < n; i++) {
          if (ara[i] == 1) cout << ara[i] << " ";
        }
        cout << k << endl;
      } else {
        if (m > 1) {
          cout << m << endl;
          for (i = 0; i < m; i++) cout << "1 ";
          cout << endl;
        } else {
          int flag = 0;
          for (i = 0; i < n; i++) {
            for (j = i + 1; j < n; j++) {
              if (prime[ara[i] + ara[j]] && !flag) {
                flag = 1;
                cout << "2\n" << ara[i] << " " << ara[j] << endl;
              }
            }
          }
          if (flag == 0)
            cout << "1\n"
                 << "1\n";
        }
      }
    } else {
      int flag = 0;
      for (i = 0; i < n; i++) {
        for (j = i + 1; j < n; j++) {
          if (prime[ara[i] + ara[j]] && !flag) {
            flag = 1;
            cout << "2\n" << ara[i] << " " << ara[j] << endl;
          }
        }
      }
      if (flag == 0) cout << "1\n" << ara[0] << endl;
    }
  }
  return 0;
}

### 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[3000000];
void precalc() {
  for (int i = 1; i < 3000000; i++) prime[i] = true;
  for (int i = 2; i < 3000000; i += 2) prime[i] = false;
  int p = sqrt(3000000);
  prime[2] = true;
  for (int i = 3; i <= p; i += 2) {
    for (int j = i * i; j < 3000000; j += 2 * i) {
      prime[j] = false;
    }
  }
}
int main() {
  int i, j, k, l, m, n, rem[1001];
  precalc();
  while (cin >> n) {
    int ara[n];
    m = 0;
    for (i = 0; i < n; i++) {
      cin >> ara[i];
      if (ara[i] == 1) m++;
    }
    if (m) {
      k = 0;
      for (i = 0; i < n; i++) {
        if (ara[i] != 1 && prime[ara[i] + 1]) {
          k = ara[i];
        }
      }
      if (k) {
        cout << m + 1 << endl;
        for (i = 0; i < n; i++) {
          if (ara[i] == 1) cout << ara[i] << " ";
        }
        cout << k << endl;
      } else {
        if (m > 1) {
          cout << m << endl;
          for (i = 0; i < m; i++) cout << "1 ";
          cout << endl;
        } else {
          int flag = 0;
          for (i = 0; i < n; i++) {
            for (j = i + 1; j < n; j++) {
              if (prime[ara[i] + ara[j]] && !flag) {
                flag = 1;
                cout << "2\n" << ara[i] << " " << ara[j] << endl;
              }
            }
          }
          if (flag == 0)
            cout << "1\n"
                 << "1\n";
        }
      }
    } else {
      int flag = 0;
      for (i = 0; i < n; i++) {
        for (j = i + 1; j < n; j++) {
          if (prime[ara[i] + ara[j]] && !flag) {
            flag = 1;
            cout << "2\n" << ara[i] << " " << ara[j] << endl;
          }
        }
      }
      if (flag == 0) cout << "1\n" << ara[0] << endl;
    }
  }
  return 0;
}

```