#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1005, MAXP = 2000005;
int n, ones, p1 = -1, p2 = -1, alt, arr[MAXN];
bool prime[MAXP];
vector<int> graph[MAXN];
int main() {
  memset(prime, 1, sizeof(prime));
  for (int i = 2; i <= 1500; i++) {
    if (prime[i]) {
      for (int j = i * i; j < MAXP; j += i) prime[j] = false;
    }
  }
  scanf("%i", &n);
  for (int i = 0; i < n; i++) scanf("%i", &arr[i]);
  for (int i = 0; i < n; i++) {
    ones += arr[i] == 1;
    for (int j = i + 1; j < n; j++) {
      if (prime[arr[i] + arr[j]]) {
        if (arr[i] == 1 && arr[j] != 1)
          alt = arr[j];
        else if (arr[i] != 1 && arr[j] == 1)
          alt = arr[i];
        if (p1 == -1) {
          p1 = arr[i];
          p2 = arr[j];
        }
      }
    }
  }
  if (ones + !!alt > 2) {
    printf("%i\n", ones + !!alt);
    for (int i = 0; i < ones; i++) printf("1 ");
    if (alt) printf("%i\n", alt);
  } else if (p1 != -1) {
    printf("2\n%i %i\n", p1, p2);
  } else {
    printf("1\n%i\n", arr[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;
const int MAXN = 1005, MAXP = 2000005;
int n, ones, p1 = -1, p2 = -1, alt, arr[MAXN];
bool prime[MAXP];
vector<int> graph[MAXN];
int main() {
  memset(prime, 1, sizeof(prime));
  for (int i = 2; i <= 1500; i++) {
    if (prime[i]) {
      for (int j = i * i; j < MAXP; j += i) prime[j] = false;
    }
  }
  scanf("%i", &n);
  for (int i = 0; i < n; i++) scanf("%i", &arr[i]);
  for (int i = 0; i < n; i++) {
    ones += arr[i] == 1;
    for (int j = i + 1; j < n; j++) {
      if (prime[arr[i] + arr[j]]) {
        if (arr[i] == 1 && arr[j] != 1)
          alt = arr[j];
        else if (arr[i] != 1 && arr[j] == 1)
          alt = arr[i];
        if (p1 == -1) {
          p1 = arr[i];
          p2 = arr[j];
        }
      }
    }
  }
  if (ones + !!alt > 2) {
    printf("%i\n", ones + !!alt);
    for (int i = 0; i < ones; i++) printf("1 ");
    if (alt) printf("%i\n", alt);
  } else if (p1 != -1) {
    printf("2\n%i %i\n", p1, p2);
  } else {
    printf("1\n%i\n", arr[0]);
  }
}

```