#include <bits/stdc++.h>
using namespace std;
const int N = (int)2e6 + 11;
int a[N];
int a1, a2;
bool cmpo[N];
void s(void) {
  cmpo[1] = 1;
  for (int i = 2; i < sqrt(N); i++) {
    if (!cmpo[i]) {
      for (int j = 2 * i; j < N; j += i) cmpo[j] = 1;
    }
  }
}
int main() {
  s();
  int n;
  cin >> n;
  int ctone = 0;
  int flag2 = 0;
  int flag1 = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1)
      ctone++;
    else if (a[i] == 2)
      flag2 = 1;
  }
  int ans1 = ctone * flag2;
  int p = 0;
  for (int i = 0; i < n; i++) {
    if (!cmpo[1 + a[i]] && ((a[i] & 1) == 0)) {
      p = a[i];
    }
    for (int j = i + 1; j < n; j++) {
      if (!cmpo[a[i] + a[j]]) {
        if (((a[i] & 1) && !(a[j] & 1)) || ((a[j] & 1) && !(a[i] & 1))) {
          flag1 = 1;
          a1 = a[i];
          a2 = a[j];
        }
      }
    }
  }
  if (ans1 + 1 >= 2) {
    cout << ctone + 1 << endl;
    for (int i = 0; i < ctone; i++) cout << 1 << " ";
    cout << 2 << endl;
    return 0;
  } else if (ctone >= 2 && p) {
    cout << ctone + 1 << endl;
    for (int i = 0; i < ctone; i++) cout << 1 << " ";
    cout << p << endl;
  } else if (ctone >= 2) {
    cout << ctone << endl;
    for (int i = 0; i < ctone; i++) cout << 1 << " ";
    cout << endl;
  } else if (flag1) {
    cout << 2 << endl;
    cout << a1 << " " << a2 << endl;
  } else
    cout << 1 << "\n" << a[0] << endl;
  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;
const int N = (int)2e6 + 11;
int a[N];
int a1, a2;
bool cmpo[N];
void s(void) {
  cmpo[1] = 1;
  for (int i = 2; i < sqrt(N); i++) {
    if (!cmpo[i]) {
      for (int j = 2 * i; j < N; j += i) cmpo[j] = 1;
    }
  }
}
int main() {
  s();
  int n;
  cin >> n;
  int ctone = 0;
  int flag2 = 0;
  int flag1 = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1)
      ctone++;
    else if (a[i] == 2)
      flag2 = 1;
  }
  int ans1 = ctone * flag2;
  int p = 0;
  for (int i = 0; i < n; i++) {
    if (!cmpo[1 + a[i]] && ((a[i] & 1) == 0)) {
      p = a[i];
    }
    for (int j = i + 1; j < n; j++) {
      if (!cmpo[a[i] + a[j]]) {
        if (((a[i] & 1) && !(a[j] & 1)) || ((a[j] & 1) && !(a[i] & 1))) {
          flag1 = 1;
          a1 = a[i];
          a2 = a[j];
        }
      }
    }
  }
  if (ans1 + 1 >= 2) {
    cout << ctone + 1 << endl;
    for (int i = 0; i < ctone; i++) cout << 1 << " ";
    cout << 2 << endl;
    return 0;
  } else if (ctone >= 2 && p) {
    cout << ctone + 1 << endl;
    for (int i = 0; i < ctone; i++) cout << 1 << " ";
    cout << p << endl;
  } else if (ctone >= 2) {
    cout << ctone << endl;
    for (int i = 0; i < ctone; i++) cout << 1 << " ";
    cout << endl;
  } else if (flag1) {
    cout << 2 << endl;
    cout << a1 << " " << a2 << endl;
  } else
    cout << 1 << "\n" << a[0] << endl;
  return 0;
}

```