#include <bits/stdc++.h>
using namespace std;
const int maxn = 1234;
int n, a[maxn], freq[1000007], prime[2000007];
void siev() {
  for (int i = 2; i * i <= 2000005; i++) {
    if (!prime[i]) continue;
    for (int j = i + i; j <= 2000005; j += i) {
      prime[j] = false;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  for (int i = 0; i < 2000007; i++) prime[i] = true;
  siev();
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    freq[a[i]]++;
  }
  if (freq[1] > 1) {
    for (int i = 0; i < n; i++)
      if (prime[a[i] + 1] && a[i] != 1) {
        cout << freq[1] + 1 << '\n';
        for (int j = 0; j < freq[1]; j++) cout << 1 << " ";
        cout << a[i];
        return 0;
      }
    cout << freq[1] << '\n';
    for (int j = 0; j < freq[1]; j++) cout << 1 << " ";
    return 0;
  }
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++)
      if (prime[a[i] + a[j]]) {
        cout << 2 << '\n' << a[i] << " " << a[j];
        return 0;
      }
  cout << 1 << '\n' << a[0];
  return 0;
}

### Prompt

In CPP, your task is to solve 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 maxn = 1234;
int n, a[maxn], freq[1000007], prime[2000007];
void siev() {
  for (int i = 2; i * i <= 2000005; i++) {
    if (!prime[i]) continue;
    for (int j = i + i; j <= 2000005; j += i) {
      prime[j] = false;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  for (int i = 0; i < 2000007; i++) prime[i] = true;
  siev();
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    freq[a[i]]++;
  }
  if (freq[1] > 1) {
    for (int i = 0; i < n; i++)
      if (prime[a[i] + 1] && a[i] != 1) {
        cout << freq[1] + 1 << '\n';
        for (int j = 0; j < freq[1]; j++) cout << 1 << " ";
        cout << a[i];
        return 0;
      }
    cout << freq[1] << '\n';
    for (int j = 0; j < freq[1]; j++) cout << 1 << " ";
    return 0;
  }
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++)
      if (prime[a[i] + a[j]]) {
        cout << 2 << '\n' << a[i] << " " << a[j];
        return 0;
      }
  cout << 1 << '\n' << a[0];
  return 0;
}

```