#include <bits/stdc++.h>
using namespace std;
vector<bool> bs;
void sieve(int MAX) {
  bs.assign(MAX, true);
  bs[0] = bs[1] = false;
  for (int i = 2; i < MAX; i++) {
    if (!bs[i]) continue;
    for (int j = 2 * i; j < MAX; j += i) bs[j] = false;
  }
}
int main(void) {
  sieve(2e6 + 10);
  int n, i;
  scanf("%d", &n);
  int a[n];
  for (i = 0; i < n; i++) scanf("%d", a + i);
  sort(a, a + n);
  vector<int> ans;
  for (i = 0; i < n && a[i] == 1; i++) ans.push_back(1);
  if (ans.size() > 1) {
    for (; i < n; i++) {
      if (bs[1 + a[i]]) {
        ans.push_back(a[i]);
        break;
      }
    }
  } else {
    ans.clear();
    for (i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        if (bs[a[i] + a[j]]) {
          ans.push_back(a[i]), ans.push_back(a[j]);
          j = n, i = n;
        }
      }
    }
    if (ans.size() == 0) ans.push_back(a[0]);
  }
  printf("%d\n", (int)ans.size());
  for (int x : ans) printf("%d ", x);
  return (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;
vector<bool> bs;
void sieve(int MAX) {
  bs.assign(MAX, true);
  bs[0] = bs[1] = false;
  for (int i = 2; i < MAX; i++) {
    if (!bs[i]) continue;
    for (int j = 2 * i; j < MAX; j += i) bs[j] = false;
  }
}
int main(void) {
  sieve(2e6 + 10);
  int n, i;
  scanf("%d", &n);
  int a[n];
  for (i = 0; i < n; i++) scanf("%d", a + i);
  sort(a, a + n);
  vector<int> ans;
  for (i = 0; i < n && a[i] == 1; i++) ans.push_back(1);
  if (ans.size() > 1) {
    for (; i < n; i++) {
      if (bs[1 + a[i]]) {
        ans.push_back(a[i]);
        break;
      }
    }
  } else {
    ans.clear();
    for (i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        if (bs[a[i] + a[j]]) {
          ans.push_back(a[i]), ans.push_back(a[j]);
          j = n, i = n;
        }
      }
    }
    if (ans.size() == 0) ans.push_back(a[0]);
  }
  printf("%d\n", (int)ans.size());
  for (int x : ans) printf("%d ", x);
  return (0);
}

```