#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2000001;
bool prime[MAXN];
int x[1000];
int cnt;
int main() {
  vector<int> ans;
  int ans_sz = 0;
  memset(prime, 1, sizeof prime);
  for (int i = 2; i * i <= MAXN; i++) {
    if (prime[i]) {
      for (int j = i * i; j <= MAXN; j += i) {
        prime[j] = 0;
      }
    }
  }
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x[i];
    cnt += (x[i] == 1);
  }
  ans_sz = 1;
  ans.push_back(x[0]);
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (prime[x[i] + x[j]]) {
        if (x[i] == 1 ^ x[j] == 1) {
          if (cnt + 1 > ans_sz) {
            ans_sz = cnt + 1;
            ans.clear();
            ans.push_back(x[i]);
            ans.push_back(x[j]);
            for (int i = 0; i < cnt - 1; i++) {
              ans.push_back(1);
            }
          }
        } else if (x[i] == 1 && x[j] == 1) {
          if (cnt > ans_sz) {
            ans_sz = cnt;
            ans.clear();
            for (int i = 0; i < ans_sz; i++) {
              ans.push_back(1);
            }
          }
        } else {
          if (2 > ans_sz) {
            ans_sz = 2;
            ans.clear();
            ans.push_back(x[i]);
            ans.push_back(x[j]);
          }
        }
      }
    }
  }
  cout << ans_sz << endl;
  for (int i = 0; i < ans_sz; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp to 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 = 2000001;
bool prime[MAXN];
int x[1000];
int cnt;
int main() {
  vector<int> ans;
  int ans_sz = 0;
  memset(prime, 1, sizeof prime);
  for (int i = 2; i * i <= MAXN; i++) {
    if (prime[i]) {
      for (int j = i * i; j <= MAXN; j += i) {
        prime[j] = 0;
      }
    }
  }
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> x[i];
    cnt += (x[i] == 1);
  }
  ans_sz = 1;
  ans.push_back(x[0]);
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (prime[x[i] + x[j]]) {
        if (x[i] == 1 ^ x[j] == 1) {
          if (cnt + 1 > ans_sz) {
            ans_sz = cnt + 1;
            ans.clear();
            ans.push_back(x[i]);
            ans.push_back(x[j]);
            for (int i = 0; i < cnt - 1; i++) {
              ans.push_back(1);
            }
          }
        } else if (x[i] == 1 && x[j] == 1) {
          if (cnt > ans_sz) {
            ans_sz = cnt;
            ans.clear();
            for (int i = 0; i < ans_sz; i++) {
              ans.push_back(1);
            }
          }
        } else {
          if (2 > ans_sz) {
            ans_sz = 2;
            ans.clear();
            ans.push_back(x[i]);
            ans.push_back(x[j]);
          }
        }
      }
    }
  }
  cout << ans_sz << endl;
  for (int i = 0; i < ans_sz; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}

```