#include <bits/stdc++.h>
using namespace std;
struct __s {
  __s() {
    if (1) {
      ios_base::Init i;
      cin.sync_with_stdio(0);
      cin.tie(0);
    }
  }
  ~__s() {
    if (!1)
      fprintf(stderr, "Execution time: %.3lf s.\n",
              (double)clock() / CLOCKS_PER_SEC);
    long long n;
    cin >> n;
  }
} __S;
bool p[2111111];
long long a[1111];
int main(void) {
  p[1] = true;
  for (long long i = 2; i <= 2000000; i++) {
    if (p[i]) continue;
    for (long long j = i * i; j <= 2000000; j += i) {
      p[j] = true;
    }
  }
  long long n;
  cin >> n;
  long long a1 = 0;
  long long a2 = 0;
  for (long long i = 0; i < (long long)(n); i++) {
    cin >> a[i];
    if (a[i] == 1)
      a1++;
    else if (!p[1 + a[i]])
      a2 = a[i];
  }
  if (a1 + !!a2 >= 2) {
    cout << a1 + !!a2 << '\n';
    for (long long i = 0; i < (long long)(a1); i++) cout << 1 << " ";
    if (a2) cout << a2;
    cout << '\n';
    return 0;
  }
  for (long long i = 0; i < (long long)(n); i++) {
    for (long long j = i + 1; j < n; j++) {
      if (!p[a[i] + a[j]]) {
        cout << 2 << '\n';
        cout << a[i] << " " << a[j] << '\n';
        return 0;
      }
    }
  }
  cout << 1 << '\n';
  cout << a[0] << '\n';
  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;
struct __s {
  __s() {
    if (1) {
      ios_base::Init i;
      cin.sync_with_stdio(0);
      cin.tie(0);
    }
  }
  ~__s() {
    if (!1)
      fprintf(stderr, "Execution time: %.3lf s.\n",
              (double)clock() / CLOCKS_PER_SEC);
    long long n;
    cin >> n;
  }
} __S;
bool p[2111111];
long long a[1111];
int main(void) {
  p[1] = true;
  for (long long i = 2; i <= 2000000; i++) {
    if (p[i]) continue;
    for (long long j = i * i; j <= 2000000; j += i) {
      p[j] = true;
    }
  }
  long long n;
  cin >> n;
  long long a1 = 0;
  long long a2 = 0;
  for (long long i = 0; i < (long long)(n); i++) {
    cin >> a[i];
    if (a[i] == 1)
      a1++;
    else if (!p[1 + a[i]])
      a2 = a[i];
  }
  if (a1 + !!a2 >= 2) {
    cout << a1 + !!a2 << '\n';
    for (long long i = 0; i < (long long)(a1); i++) cout << 1 << " ";
    if (a2) cout << a2;
    cout << '\n';
    return 0;
  }
  for (long long i = 0; i < (long long)(n); i++) {
    for (long long j = i + 1; j < n; j++) {
      if (!p[a[i] + a[j]]) {
        cout << 2 << '\n';
        cout << a[i] << " " << a[j] << '\n';
        return 0;
      }
    }
  }
  cout << 1 << '\n';
  cout << a[0] << '\n';
  return 0;
}

```