#include <bits/stdc++.h>
int b[2000005] = {0}, result = -1;
int a[1005];
void init() {
  for (int ii = 2; ii <= 100000; ii++) {
    for (int jj = 2; jj * ii <= 2000005; jj++) {
      b[ii * jj] = 1;
    }
  }
}
using namespace std;
int main() {
  int n, sum = 0;
  init();
  cin >> n;
  for (int ii = 1; ii <= n; ii++) {
    cin >> a[ii];
  }
  sort(a, a + n);
  if (a[1] == 1) {
    for (int ii = 1; ii <= n; ii++) {
      if (a[ii] == 1) ++sum;
    }
    for (int ii = sum; ii <= n; ii++) {
      if (b[a[ii] + 1] == 0 && a[ii] != 1) {
        result = a[ii];
      }
    }
    if (result == -1 && sum != 1) {
      cout << sum << endl;
      for (int ii = 1; ii <= sum; ii++) cout << 1 << ' ';
    } else if (result == -1 && sum == 1) {
      for (int ii = 2; ii <= n; ii++)
        for (int jj = 2; jj <= n; jj++) {
          if (b[a[ii] + a[jj]] == 0) {
            result = 2;
            cout << 2 << endl << a[ii] << ' ' << a[jj] << endl;
            return 0;
          }
        }
      if (result == -1) {
        for (int ii = 1; ii <= n; ii++) {
          if (b[a[ii]] == 0) {
            result = 1;
            cout << 1 << endl << a[ii] << endl;
            return 0;
          }
        }
      }
      cout << 1 << endl << a[1] << endl;
    } else {
      cout << sum + 1 << endl;
      for (int ii = 1; ii <= sum; ii++) cout << 1 << ' ';
      cout << result << endl;
    }
  } else {
    for (int ii = 1; ii <= n; ii++)
      for (int jj = 1; jj <= n; jj++) {
        if (b[a[ii] + a[jj]] == 0) {
          result = 2;
          cout << 2 << endl << a[ii] << ' ' << a[jj] << endl;
          return 0;
        }
      }
    if (result == -1) {
      for (int ii = 1; ii <= n; ii++) {
        if (b[a[ii]] == 0) {
          result = 1;
          cout << 1 << endl << a[ii] << endl;
          return 0;
        }
      }
    }
    cout << 1 << endl << a[1] << endl;
  }
  return 0;
}

### Prompt

Please formulate 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>
int b[2000005] = {0}, result = -1;
int a[1005];
void init() {
  for (int ii = 2; ii <= 100000; ii++) {
    for (int jj = 2; jj * ii <= 2000005; jj++) {
      b[ii * jj] = 1;
    }
  }
}
using namespace std;
int main() {
  int n, sum = 0;
  init();
  cin >> n;
  for (int ii = 1; ii <= n; ii++) {
    cin >> a[ii];
  }
  sort(a, a + n);
  if (a[1] == 1) {
    for (int ii = 1; ii <= n; ii++) {
      if (a[ii] == 1) ++sum;
    }
    for (int ii = sum; ii <= n; ii++) {
      if (b[a[ii] + 1] == 0 && a[ii] != 1) {
        result = a[ii];
      }
    }
    if (result == -1 && sum != 1) {
      cout << sum << endl;
      for (int ii = 1; ii <= sum; ii++) cout << 1 << ' ';
    } else if (result == -1 && sum == 1) {
      for (int ii = 2; ii <= n; ii++)
        for (int jj = 2; jj <= n; jj++) {
          if (b[a[ii] + a[jj]] == 0) {
            result = 2;
            cout << 2 << endl << a[ii] << ' ' << a[jj] << endl;
            return 0;
          }
        }
      if (result == -1) {
        for (int ii = 1; ii <= n; ii++) {
          if (b[a[ii]] == 0) {
            result = 1;
            cout << 1 << endl << a[ii] << endl;
            return 0;
          }
        }
      }
      cout << 1 << endl << a[1] << endl;
    } else {
      cout << sum + 1 << endl;
      for (int ii = 1; ii <= sum; ii++) cout << 1 << ' ';
      cout << result << endl;
    }
  } else {
    for (int ii = 1; ii <= n; ii++)
      for (int jj = 1; jj <= n; jj++) {
        if (b[a[ii] + a[jj]] == 0) {
          result = 2;
          cout << 2 << endl << a[ii] << ' ' << a[jj] << endl;
          return 0;
        }
      }
    if (result == -1) {
      for (int ii = 1; ii <= n; ii++) {
        if (b[a[ii]] == 0) {
          result = 1;
          cout << 1 << endl << a[ii] << endl;
          return 0;
        }
      }
    }
    cout << 1 << endl << a[1] << endl;
  }
  return 0;
}

```