#include <bits/stdc++.h>
using namespace std;
int n, a[1002], L, R, cnt, mx, M, b[1002], k;
bool prime[2000003];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == 1) cnt++;
  }
  for (int i = 2; i <= (int)2e6; i++) prime[i] = true;
  for (int i = 2; i <= (int)2e6; i++)
    if (prime[i] && i * 1ll * i <= (int)2e6)
      for (int j = i * 1ll * i; j <= (int)2e6; j += i) prime[j] = false;
  for (int i = 1; i <= n; i++)
    if (a[i] != 1) b[++k] = a[i];
  if (cnt)
    for (int i = 1; i <= k; i++)
      if (prime[1 + b[i]]) M = i;
  for (int i = 1; i <= k; i++)
    for (int j = i + 1; j <= k; j++) {
      if (b[i] != b[j] && prime[b[i] + b[j]]) L = i, R = j;
    }
  if (cnt == 0 && L == 0) {
    cout << 1 << endl << a[1];
  } else {
    int cur = cnt + (M != 0);
    if (cur >= 2) {
      cout << cur << endl;
      for (int i = 0; i < cnt; i++) cout << 1 << ' ';
      if (M != 0) cout << b[M];
    } else {
      if (L == 0)
        cout << 1 << endl << a[1] << endl;
      else
        cout << 2 << endl << b[L] << ' ' << b[R];
    }
  }
  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;
int n, a[1002], L, R, cnt, mx, M, b[1002], k;
bool prime[2000003];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] == 1) cnt++;
  }
  for (int i = 2; i <= (int)2e6; i++) prime[i] = true;
  for (int i = 2; i <= (int)2e6; i++)
    if (prime[i] && i * 1ll * i <= (int)2e6)
      for (int j = i * 1ll * i; j <= (int)2e6; j += i) prime[j] = false;
  for (int i = 1; i <= n; i++)
    if (a[i] != 1) b[++k] = a[i];
  if (cnt)
    for (int i = 1; i <= k; i++)
      if (prime[1 + b[i]]) M = i;
  for (int i = 1; i <= k; i++)
    for (int j = i + 1; j <= k; j++) {
      if (b[i] != b[j] && prime[b[i] + b[j]]) L = i, R = j;
    }
  if (cnt == 0 && L == 0) {
    cout << 1 << endl << a[1];
  } else {
    int cur = cnt + (M != 0);
    if (cur >= 2) {
      cout << cur << endl;
      for (int i = 0; i < cnt; i++) cout << 1 << ' ';
      if (M != 0) cout << b[M];
    } else {
      if (L == 0)
        cout << 1 << endl << a[1] << endl;
      else
        cout << 2 << endl << b[L] << ' ' << b[R];
    }
  }
  return 0;
}

```