#include <bits/stdc++.h>
using namespace std;
vector<bool> sieveFlags(int n) {
  vector<bool> f(n + 1);
  for (int i = 3; i <= n; i += 2) f[i] = 1;
  for (int i = 3; i <= sqrt(n); i += 2)
    if (f[i])
      for (int j = i * 3; j <= n; j += i) f[j] = 0;
  f[2] = 1;
  return f;
}
int n, a[1001], vis[1001], color[1001];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < (int)n; ++i) cin >> a[i];
  auto isP = sieveFlags(3 * 1000000);
  int oneCount = count(a, a + n, 1);
  vector<int> res;
  if (oneCount >= 2) {
    res = vector<int>(oneCount, 1);
    for (int i = 0; i < (int)n; ++i)
      if (isP[1 + a[i]] && a[i] != 1) {
        res.push_back(a[i]);
        break;
      }
  } else {
    [&]() {
      for (int i = 0; i < (int)n; ++i)
        for (int j = i + 1; j < (int)n; ++j)
          if (isP[a[i] + a[j]]) {
            res.push_back(a[i]);
            res.push_back(a[j]);
            return;
          }
    }();
    if ((int)(res).size() == 0) res.push_back(a[0]);
  }
  int m = (int)(res).size();
  cout << m << endl;
  for (int i = 0; i < (int)m; ++i) cout << res[i] << (i == m - 1 ? '\n' : ' ');
}

### 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;
vector<bool> sieveFlags(int n) {
  vector<bool> f(n + 1);
  for (int i = 3; i <= n; i += 2) f[i] = 1;
  for (int i = 3; i <= sqrt(n); i += 2)
    if (f[i])
      for (int j = i * 3; j <= n; j += i) f[j] = 0;
  f[2] = 1;
  return f;
}
int n, a[1001], vis[1001], color[1001];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < (int)n; ++i) cin >> a[i];
  auto isP = sieveFlags(3 * 1000000);
  int oneCount = count(a, a + n, 1);
  vector<int> res;
  if (oneCount >= 2) {
    res = vector<int>(oneCount, 1);
    for (int i = 0; i < (int)n; ++i)
      if (isP[1 + a[i]] && a[i] != 1) {
        res.push_back(a[i]);
        break;
      }
  } else {
    [&]() {
      for (int i = 0; i < (int)n; ++i)
        for (int j = i + 1; j < (int)n; ++j)
          if (isP[a[i] + a[j]]) {
            res.push_back(a[i]);
            res.push_back(a[j]);
            return;
          }
    }();
    if ((int)(res).size() == 0) res.push_back(a[0]);
  }
  int m = (int)(res).size();
  cout << m << endl;
  for (int i = 0; i < (int)m; ++i) cout << res[i] << (i == m - 1 ? '\n' : ' ');
}

```