#include <bits/stdc++.h>
using namespace std;
const int N = 1000 + 5;
const int M = N * N * 2;
int n;
int a[N], is_prime[M];
void sieve() {
  for (int(i) = (0); (i) < (int)(M); (i)++) is_prime[i] = 1;
  is_prime[0] = is_prime[1] = 0;
  for (int(i) = (2); (i) < (int)(M); (i)++) {
    if (is_prime[i]) {
      for (int j = i * 2; j < M; j += i) is_prime[j] = 0;
    }
  }
}
int main() {
  sieve();
  cin >> n;
  for (int(i) = (0); (i) < (int)(n); (i)++) cin >> a[i];
  sort(a, a + n);
  if (a[0] == 1 && n > 1 && a[1] == 1) {
    int pos = upper_bound(a, a + n, a[0]) - a;
    int num = -1;
    for (int(i) = (pos); (i) < (int)(n); (i)++)
      if (is_prime[a[i] + 1]) {
        num = a[i];
        break;
      }
    if (num == -1) {
      cout << pos << endl;
      while (pos--) cout << 1 << " ";
      cout << endl;
    } else {
      cout << pos + 1 << endl;
      while (pos--) cout << 1 << " ";
      cout << num << endl;
    }
  } else {
    bool ok = 0;
    for (int(i) = (0); (i) < (int)(n); (i)++) {
      for (int(j) = (i + 1); (j) < (int)(n); (j)++)
        if (is_prime[a[i] + a[j]]) {
          cout << 2 << endl;
          cout << a[i] << " " << a[j] << endl;
          ok = 1;
          break;
        }
      if (ok) break;
    }
    if (!ok) {
      cout << 1 << endl << a[0] << 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>
using namespace std;
const int N = 1000 + 5;
const int M = N * N * 2;
int n;
int a[N], is_prime[M];
void sieve() {
  for (int(i) = (0); (i) < (int)(M); (i)++) is_prime[i] = 1;
  is_prime[0] = is_prime[1] = 0;
  for (int(i) = (2); (i) < (int)(M); (i)++) {
    if (is_prime[i]) {
      for (int j = i * 2; j < M; j += i) is_prime[j] = 0;
    }
  }
}
int main() {
  sieve();
  cin >> n;
  for (int(i) = (0); (i) < (int)(n); (i)++) cin >> a[i];
  sort(a, a + n);
  if (a[0] == 1 && n > 1 && a[1] == 1) {
    int pos = upper_bound(a, a + n, a[0]) - a;
    int num = -1;
    for (int(i) = (pos); (i) < (int)(n); (i)++)
      if (is_prime[a[i] + 1]) {
        num = a[i];
        break;
      }
    if (num == -1) {
      cout << pos << endl;
      while (pos--) cout << 1 << " ";
      cout << endl;
    } else {
      cout << pos + 1 << endl;
      while (pos--) cout << 1 << " ";
      cout << num << endl;
    }
  } else {
    bool ok = 0;
    for (int(i) = (0); (i) < (int)(n); (i)++) {
      for (int(j) = (i + 1); (j) < (int)(n); (j)++)
        if (is_prime[a[i] + a[j]]) {
          cout << 2 << endl;
          cout << a[i] << " " << a[j] << endl;
          ok = 1;
          break;
        }
      if (ok) break;
    }
    if (!ok) {
      cout << 1 << endl << a[0] << endl;
    }
  }
  return 0;
}

```