#include <bits/stdc++.h>
using namespace std;
int a[2000001];
vector<int> v;
bool prime(int x) {
  for (int i = 2; i * i <= x; ++i) {
    if (x % i == 0) return 0;
  }
  return 1;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  sort(a, a + n);
  bool yes = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = n - 1; j > i; --j) {
      if (prime(a[i] + a[j])) {
        if (a[i] == 1) {
          if (a[j] != 1) v.push_back(a[j]);
          yes = 1;
          break;
        } else {
          puts("2");
          printf("%d %d\n", a[i], a[j]);
          return 0;
        }
      }
    }
    if (yes) break;
  }
  if (yes) {
    for (int i = 0; i < n; ++i) {
      if (a[i] == 1) v.push_back(a[i]);
    }
    printf("%d\n", (int)v.size());
    for (int i = 0; i < v.size(); ++i) printf("%d ", v[i]);
  } else {
    printf("1\n%d", a[0]);
  }
  return 0;
}

### Prompt

Generate 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;
int a[2000001];
vector<int> v;
bool prime(int x) {
  for (int i = 2; i * i <= x; ++i) {
    if (x % i == 0) return 0;
  }
  return 1;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  sort(a, a + n);
  bool yes = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = n - 1; j > i; --j) {
      if (prime(a[i] + a[j])) {
        if (a[i] == 1) {
          if (a[j] != 1) v.push_back(a[j]);
          yes = 1;
          break;
        } else {
          puts("2");
          printf("%d %d\n", a[i], a[j]);
          return 0;
        }
      }
    }
    if (yes) break;
  }
  if (yes) {
    for (int i = 0; i < n; ++i) {
      if (a[i] == 1) v.push_back(a[i]);
    }
    printf("%d\n", (int)v.size());
    for (int i = 0; i < v.size(); ++i) printf("%d ", v[i]);
  } else {
    printf("1\n%d", a[0]);
  }
  return 0;
}

```