#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2000005;
int prime[MAXN];
int main() {
  int N, ones = 0, twos = 0;
  scanf("%d", &N);
  int arr[N];
  vector<int> primes;
  for (int i = 2; i < MAXN; i++) {
    if (!prime[i]) {
      primes.push_back(i);
      for (int j = i + i; j < MAXN; j += i) {
        prime[j] = 1;
      }
    }
  }
  for (int i = 0; i < N; i++) {
    scanf("%d", &arr[i]);
    if (arr[i] == 1) ones++;
  }
  int ans = 0;
  int res = -1;
  for (int i = 0; i < N; i++) {
    if (arr[i] % 2 == 0) {
      if (!prime[arr[i] + 1]) {
        res = i;
      }
    }
  }
  if (res != -1 && ones >= 1) {
    printf("%d\n", ones + 1);
    for (int i = 0; i < ones; i++) {
      printf("1 ");
    }
    printf("%d\n", arr[res]);
    return 0;
  }
  if (ones >= 2) {
    printf("%d\n", ones);
    for (int i = 0; i < ones; i++) {
      printf("1 ");
    }
    return 0;
  }
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      if (i == j) continue;
      if (arr[i] % 2 == 0 && !prime[arr[i] + arr[j]]) {
        printf("2\n%d %d\n", arr[i], arr[j]);
        return 0;
      }
    }
  }
  printf("1\n%d\n", arr[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 MAXN = 2000005;
int prime[MAXN];
int main() {
  int N, ones = 0, twos = 0;
  scanf("%d", &N);
  int arr[N];
  vector<int> primes;
  for (int i = 2; i < MAXN; i++) {
    if (!prime[i]) {
      primes.push_back(i);
      for (int j = i + i; j < MAXN; j += i) {
        prime[j] = 1;
      }
    }
  }
  for (int i = 0; i < N; i++) {
    scanf("%d", &arr[i]);
    if (arr[i] == 1) ones++;
  }
  int ans = 0;
  int res = -1;
  for (int i = 0; i < N; i++) {
    if (arr[i] % 2 == 0) {
      if (!prime[arr[i] + 1]) {
        res = i;
      }
    }
  }
  if (res != -1 && ones >= 1) {
    printf("%d\n", ones + 1);
    for (int i = 0; i < ones; i++) {
      printf("1 ");
    }
    printf("%d\n", arr[res]);
    return 0;
  }
  if (ones >= 2) {
    printf("%d\n", ones);
    for (int i = 0; i < ones; i++) {
      printf("1 ");
    }
    return 0;
  }
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      if (i == j) continue;
      if (arr[i] % 2 == 0 && !prime[arr[i] + arr[j]]) {
        printf("2\n%d %d\n", arr[i], arr[j]);
        return 0;
      }
    }
  }
  printf("1\n%d\n", arr[0]);
}

```