#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e3 + 10;
const int MAXM = 2e6 + 10;
int n, num;
bool ok[MAXM];
int arr[MAXN];
vector<int> e;
void solve() {
  for (int i = 2; i < MAXM; i++) ok[i] = true;
  for (int i = 2; i < MAXM; i++) {
    if (ok[i]) {
      for (int j = i + i; j < MAXM; j += i) ok[j] = false;
    }
  }
}
int main() {
  scanf("%d", &n);
  solve();
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    if (arr[i] == 1) num++;
  }
  if (num < 2) {
    int flag = 0;
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++)
        if (ok[arr[i] + arr[j]]) {
          e.push_back(arr[i]);
          e.push_back(arr[j]);
          flag = 1;
          break;
        }
      if (flag) break;
    }
    if (e.empty()) {
      for (int i = 0; i < n; i++)
        if (ok[arr[i]]) {
          e.push_back(arr[i]);
          break;
        }
    }
  } else {
    int flag = 1, cnt = 0;
    for (int i = 0; i < n; i++) {
      if (arr[i] == 1)
        e.push_back(arr[i]);
      else if (ok[arr[i] + 1] && ok[arr[i] + flag]) {
        e.push_back(arr[i]);
        flag = arr[i];
        cnt++;
      }
      if (cnt == 2) break;
    }
  }
  if (e.empty())
    printf("1\n%d\n", arr[0]);
  else {
    printf("%d\n", (int)e.size());
    for (int i = 0; i < (int)e.size(); i++) {
      printf("%d ", e[i]);
    }
    printf("\n");
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 = 2e3 + 10;
const int MAXM = 2e6 + 10;
int n, num;
bool ok[MAXM];
int arr[MAXN];
vector<int> e;
void solve() {
  for (int i = 2; i < MAXM; i++) ok[i] = true;
  for (int i = 2; i < MAXM; i++) {
    if (ok[i]) {
      for (int j = i + i; j < MAXM; j += i) ok[j] = false;
    }
  }
}
int main() {
  scanf("%d", &n);
  solve();
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    if (arr[i] == 1) num++;
  }
  if (num < 2) {
    int flag = 0;
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++)
        if (ok[arr[i] + arr[j]]) {
          e.push_back(arr[i]);
          e.push_back(arr[j]);
          flag = 1;
          break;
        }
      if (flag) break;
    }
    if (e.empty()) {
      for (int i = 0; i < n; i++)
        if (ok[arr[i]]) {
          e.push_back(arr[i]);
          break;
        }
    }
  } else {
    int flag = 1, cnt = 0;
    for (int i = 0; i < n; i++) {
      if (arr[i] == 1)
        e.push_back(arr[i]);
      else if (ok[arr[i] + 1] && ok[arr[i] + flag]) {
        e.push_back(arr[i]);
        flag = arr[i];
        cnt++;
      }
      if (cnt == 2) break;
    }
  }
  if (e.empty())
    printf("1\n%d\n", arr[0]);
  else {
    printf("%d\n", (int)e.size());
    for (int i = 0; i < (int)e.size(); i++) {
      printf("%d ", e[i]);
    }
    printf("\n");
  }
  return 0;
}

```