#include <bits/stdc++.h>
using namespace std;
int cnt;
int arr[1003];
bool noprime[2000003];
void sieve() {
  noprime[0] = noprime[1] = 1;
  for (int i = 2; i <= 2000000; i++) {
    if (!noprime[i]) {
      for (int j = i; 1LL * j * i <= 2000000; j++) noprime[i * j] = 1;
    }
  }
}
int main() {
  sieve();
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    if (arr[i] == 1) cnt++;
  }
  if (cnt < 2) {
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        if (!noprime[arr[i] + arr[j]]) {
          printf("%d\n%d %d\n", 2, arr[i], arr[j]);
          return 0;
        }
      }
    }
    printf("1\n%d\n", arr[0]);
  } else {
    int x = -1;
    for (int i = 0; i < n; i++) {
      if (!noprime[arr[i] + 1] && arr[i] > 1) {
        x = arr[i];
        break;
      }
    }
    if (x == -1) {
      printf("%d\n1", cnt);
      for (int i = 1; i < cnt; i++) printf(" 1");
      printf("\n");
    } else {
      printf("%d\n%d", cnt + 1, x);
      for (int i = 1; i <= cnt; i++) printf(" 1");
      printf("\n");
    }
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
int cnt;
int arr[1003];
bool noprime[2000003];
void sieve() {
  noprime[0] = noprime[1] = 1;
  for (int i = 2; i <= 2000000; i++) {
    if (!noprime[i]) {
      for (int j = i; 1LL * j * i <= 2000000; j++) noprime[i * j] = 1;
    }
  }
}
int main() {
  sieve();
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    if (arr[i] == 1) cnt++;
  }
  if (cnt < 2) {
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        if (!noprime[arr[i] + arr[j]]) {
          printf("%d\n%d %d\n", 2, arr[i], arr[j]);
          return 0;
        }
      }
    }
    printf("1\n%d\n", arr[0]);
  } else {
    int x = -1;
    for (int i = 0; i < n; i++) {
      if (!noprime[arr[i] + 1] && arr[i] > 1) {
        x = arr[i];
        break;
      }
    }
    if (x == -1) {
      printf("%d\n1", cnt);
      for (int i = 1; i < cnt; i++) printf(" 1");
      printf("\n");
    } else {
      printf("%d\n%d", cnt + 1, x);
      for (int i = 1; i <= cnt; i++) printf(" 1");
      printf("\n");
    }
  }
}

```