#include <bits/stdc++.h>
using namespace std;
const int N = 1000010;
bool composite[2 * N];
int has[N];
int n;
vector<int> v;
int main() {
  for (int i = 2; i * i < 2 * N; i++) {
    if (!composite[i]) {
      for (int j = i * i; j < 2 * N; j += i) {
        composite[j] = true;
      }
    }
  }
  vector<int> ans;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    if (!has[x]) {
      v.push_back(x);
    }
    has[x]++;
  }
  if (has[1] > 1) {
    for (int i = 0; i < has[1]; i++) ans.push_back(1);
    bool ok = false;
    for (int i = 0; i < v.size() and !ok; i++) {
      if (v[i] == 1) continue;
      if (!composite[v[i] + 1]) {
        ans.push_back(v[i]);
        ok = true;
      }
    }
  } else {
    bool ok = false;
    for (int i = 0; i < v.size() and !ok; i++) {
      for (int j = i + 1; j < v.size() and !ok; j++) {
        if (!composite[v[i] + v[j]]) {
          ans.push_back(v[i]);
          ans.push_back(v[j]);
          ok = true;
        }
      }
    }
    if (!ok) ans.push_back(v[0]);
  }
  printf("%d\n", (int)ans.size());
  for (int i = 0; i < ans.size(); i++) {
    if (i) printf(" ");
    printf("%d", ans[i]);
  }
  printf("\n");
  return 0;
}

### Prompt

Please formulate a Cpp solution 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 N = 1000010;
bool composite[2 * N];
int has[N];
int n;
vector<int> v;
int main() {
  for (int i = 2; i * i < 2 * N; i++) {
    if (!composite[i]) {
      for (int j = i * i; j < 2 * N; j += i) {
        composite[j] = true;
      }
    }
  }
  vector<int> ans;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    if (!has[x]) {
      v.push_back(x);
    }
    has[x]++;
  }
  if (has[1] > 1) {
    for (int i = 0; i < has[1]; i++) ans.push_back(1);
    bool ok = false;
    for (int i = 0; i < v.size() and !ok; i++) {
      if (v[i] == 1) continue;
      if (!composite[v[i] + 1]) {
        ans.push_back(v[i]);
        ok = true;
      }
    }
  } else {
    bool ok = false;
    for (int i = 0; i < v.size() and !ok; i++) {
      for (int j = i + 1; j < v.size() and !ok; j++) {
        if (!composite[v[i] + v[j]]) {
          ans.push_back(v[i]);
          ans.push_back(v[j]);
          ok = true;
        }
      }
    }
    if (!ok) ans.push_back(v[0]);
  }
  printf("%d\n", (int)ans.size());
  for (int i = 0; i < ans.size(); i++) {
    if (i) printf(" ");
    printf("%d", ans[i]);
  }
  printf("\n");
  return 0;
}

```