#include <bits/stdc++.h>
using namespace std;
const int MAX = 2000100;
bool p[MAX];
void sieve() {
  memset(p, -1, sizeof p);
  p[0] = p[1] = 0;
  int s = (int)sqrt((double)MAX);
  for (int i = 2; i <= s; ++i)
    if (p[i])
      for (int j = i * i; j < MAX; j += i) p[j] = 0;
}
int main() {
  sieve();
  int n;
  cin >> n;
  int a[100100];
  for (int i = 0; i < n; i++) cin >> a[i];
  bool found = 0;
  int idx = -1;
  for (int i = 0; i < n; i++)
    if (a[i] != 1 && p[a[i] + 1]) found = 1, idx = i;
  int ones = 0;
  for (int i = 0; i < n; i++)
    if (a[i] == 1) ones++;
  int x = -1, y = -1;
  for (int i = 0; i < n - 1; i++)
    for (int j = i + 1; j < n; j++)
      if (p[a[i] + a[j]]) x = i, y = j;
  int ans1 = ones + (idx != -1);
  int ans2 = 0;
  if (x != -1) ans2 = 2;
  if (ans1 > 0 && ans1 > ans2) {
    cout << ans1 << endl;
    if (idx != -1) cout << a[idx] << " ";
    for (int i = 0; i < ones; i++) cout << 1 << " ";
  } else if (x != -1)
    cout << 2 << endl << a[x] << " " << a[y];
  else
    cout << 1 << endl << a[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;
const int MAX = 2000100;
bool p[MAX];
void sieve() {
  memset(p, -1, sizeof p);
  p[0] = p[1] = 0;
  int s = (int)sqrt((double)MAX);
  for (int i = 2; i <= s; ++i)
    if (p[i])
      for (int j = i * i; j < MAX; j += i) p[j] = 0;
}
int main() {
  sieve();
  int n;
  cin >> n;
  int a[100100];
  for (int i = 0; i < n; i++) cin >> a[i];
  bool found = 0;
  int idx = -1;
  for (int i = 0; i < n; i++)
    if (a[i] != 1 && p[a[i] + 1]) found = 1, idx = i;
  int ones = 0;
  for (int i = 0; i < n; i++)
    if (a[i] == 1) ones++;
  int x = -1, y = -1;
  for (int i = 0; i < n - 1; i++)
    for (int j = i + 1; j < n; j++)
      if (p[a[i] + a[j]]) x = i, y = j;
  int ans1 = ones + (idx != -1);
  int ans2 = 0;
  if (x != -1) ans2 = 2;
  if (ans1 > 0 && ans1 > ans2) {
    cout << ans1 << endl;
    if (idx != -1) cout << a[idx] << " ";
    for (int i = 0; i < ones; i++) cout << 1 << " ";
  } else if (x != -1)
    cout << 2 << endl << a[x] << " " << a[y];
  else
    cout << 1 << endl << a[0];
}

```