#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
long long int arr[2000009], brr[2000009];
int main() {
  int test_case = 1;
  for (int cs = 1; cs <= test_case; cs++) {
    for (int i = 2; i <= 2000009 - 9; i++) {
      if (!arr[i]) {
        for (int j = i + i; j <= 2000009 - 9; j += i) arr[j] = 1;
      }
    }
    long long int n;
    scanf("%lld", &n);
    long long int a1 = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &brr[i]);
      if (brr[i] == 1) a1++;
    }
    if (a1) {
      long long int b1 = 0;
      for (int i = 1; i <= n; i++) {
        if (brr[i] != 1 && arr[brr[i] + 1] == 0) {
          b1 = brr[i];
          break;
        }
      }
      if ((b1 + a1) < 2) {
        for (int i = 1; i <= n; i++) {
          for (int j = i + 1; j <= n; j++) {
            if (arr[brr[i] + brr[j]] == 0) {
              printf("2\n%lld %lld\n", brr[i], brr[j]);
              return 0;
            }
          }
        }
      }
      if (b1)
        printf("%lld\n", a1 + 1);
      else
        printf("%lld\n", a1);
      if (b1) printf("%lld ", b1);
      while (a1--) printf("1 ");
      printf("\n");
      return 0;
    }
    for (int i = 1; i <= n; i++) {
      for (int j = i + 1; j <= n; j++) {
        if (arr[brr[i] + brr[j]] == 0) {
          printf("2\n%lld %lld\n", brr[i], brr[j]);
          return 0;
        }
      }
    }
    printf("1\n%lld\n", brr[1]);
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
long long int arr[2000009], brr[2000009];
int main() {
  int test_case = 1;
  for (int cs = 1; cs <= test_case; cs++) {
    for (int i = 2; i <= 2000009 - 9; i++) {
      if (!arr[i]) {
        for (int j = i + i; j <= 2000009 - 9; j += i) arr[j] = 1;
      }
    }
    long long int n;
    scanf("%lld", &n);
    long long int a1 = 0;
    for (int i = 1; i <= n; i++) {
      scanf("%lld", &brr[i]);
      if (brr[i] == 1) a1++;
    }
    if (a1) {
      long long int b1 = 0;
      for (int i = 1; i <= n; i++) {
        if (brr[i] != 1 && arr[brr[i] + 1] == 0) {
          b1 = brr[i];
          break;
        }
      }
      if ((b1 + a1) < 2) {
        for (int i = 1; i <= n; i++) {
          for (int j = i + 1; j <= n; j++) {
            if (arr[brr[i] + brr[j]] == 0) {
              printf("2\n%lld %lld\n", brr[i], brr[j]);
              return 0;
            }
          }
        }
      }
      if (b1)
        printf("%lld\n", a1 + 1);
      else
        printf("%lld\n", a1);
      if (b1) printf("%lld ", b1);
      while (a1--) printf("1 ");
      printf("\n");
      return 0;
    }
    for (int i = 1; i <= n; i++) {
      for (int j = i + 1; j <= n; j++) {
        if (arr[brr[i] + brr[j]] == 0) {
          printf("2\n%lld %lld\n", brr[i], brr[j]);
          return 0;
        }
      }
    }
    printf("1\n%lld\n", brr[1]);
  }
  return 0;
}

```