#include <bits/stdc++.h>
using namespace std;
const int inf = 1000000005;
vector<int> odd, even;
int cnt[2000005];
int su[2000005];
bool judge[2000005];
int num;
void prepare() {
  memset(judge, true, sizeof(judge));
  num = 0;
  int N = 2000005;
  for (int i = 2; i <= N; i++) {
    if (judge[i]) su[num++] = i;
    for (int j = 0; j < num; j++) {
      if (i * su[j] > N) break;
      judge[i * su[j]] = false;
      if (i % su[j] == 0) break;
    }
  }
}
int main() {
  prepare();
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    if (x % 2 == 1)
      odd.push_back(x);
    else {
      cnt[x]++;
      even.push_back(x);
    }
  }
  sort(odd.begin(), odd.end());
  sort(even.begin(), even.end());
  int num1 = 0, num2 = 0;
  for (int i = 0, j = odd.size(); i < j; i++)
    if (odd[i] == 1) num1++;
  if (num1) {
    int maxval = -1, a = -1;
    for (int i = 0; i < even.size(); i++) {
      if (judge[1 + even[i]]) {
        a = even[i];
        break;
      }
    }
    if (a != -1) {
      printf("%d\n", 1 + num1);
      for (int i = 1; i <= num1; i++) printf("1 ");
      printf("%d\n", a);
      return 0;
    } else if (num1 > 1) {
      printf("%d\n", num1);
      for (int i = 1; i <= num1; i++) printf("1%c", i == num1 ? '\n' : ' ');
      return 0;
    }
  }
  for (int i = 0; i < odd.size(); i++)
    for (int j = 0; j < even.size(); j++) {
      if (judge[odd[i] + even[j]]) {
        printf("2\n%d %d\n", odd[i], even[j]);
        return 0;
      }
    }
  if (odd.size() != 0) {
    printf("1\n%d\n", odd[0]);
    return 0;
  }
  if (even.size() != 0) {
    printf("1\n%d\n", even[0]);
    return 0;
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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 inf = 1000000005;
vector<int> odd, even;
int cnt[2000005];
int su[2000005];
bool judge[2000005];
int num;
void prepare() {
  memset(judge, true, sizeof(judge));
  num = 0;
  int N = 2000005;
  for (int i = 2; i <= N; i++) {
    if (judge[i]) su[num++] = i;
    for (int j = 0; j < num; j++) {
      if (i * su[j] > N) break;
      judge[i * su[j]] = false;
      if (i % su[j] == 0) break;
    }
  }
}
int main() {
  prepare();
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    if (x % 2 == 1)
      odd.push_back(x);
    else {
      cnt[x]++;
      even.push_back(x);
    }
  }
  sort(odd.begin(), odd.end());
  sort(even.begin(), even.end());
  int num1 = 0, num2 = 0;
  for (int i = 0, j = odd.size(); i < j; i++)
    if (odd[i] == 1) num1++;
  if (num1) {
    int maxval = -1, a = -1;
    for (int i = 0; i < even.size(); i++) {
      if (judge[1 + even[i]]) {
        a = even[i];
        break;
      }
    }
    if (a != -1) {
      printf("%d\n", 1 + num1);
      for (int i = 1; i <= num1; i++) printf("1 ");
      printf("%d\n", a);
      return 0;
    } else if (num1 > 1) {
      printf("%d\n", num1);
      for (int i = 1; i <= num1; i++) printf("1%c", i == num1 ? '\n' : ' ');
      return 0;
    }
  }
  for (int i = 0; i < odd.size(); i++)
    for (int j = 0; j < even.size(); j++) {
      if (judge[odd[i] + even[j]]) {
        printf("2\n%d %d\n", odd[i], even[j]);
        return 0;
      }
    }
  if (odd.size() != 0) {
    printf("1\n%d\n", odd[0]);
    return 0;
  }
  if (even.size() != 0) {
    printf("1\n%d\n", even[0]);
    return 0;
  }
  return 0;
}

```