#include <bits/stdc++.h>
using namespace std;
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
vector<int> v;
int arr[100009], p[100009 << 5];
void yaz() {
  printf("%d\n", int(v.size()));
  for (int i = 0; i < v.size(); i++) printf("%d ", v[i]);
  printf("\n");
}
int prime(int x) {
  p[1] = 2;
  return (p[x] == 1);
}
int main() {
  for (int i = 2; i <= 2000000; i++) {
    if (p[i] == 2) continue;
    p[i] = 1;
    for (int j = i * 2; j <= 2000000; j += i) p[j] = 2;
  }
  int a, bir = 0;
  scanf("%d", &a);
  for (int i = 1; i <= a; i++) {
    scanf("%d", arr + i);
    if (arr[i] == 1) bir++;
  }
  if (bir > 0) {
    for (int i = 1; i <= a; i++)
      if (arr[i] != 1 and prime(arr[i] + 1)) {
        for (int j = 1; j <= bir; j++) v.push_back(1);
        v.push_back(arr[i]);
        yaz();
        return 0;
      }
  }
  if (bir >= 2) {
    for (int i = 1; i <= bir; i++) v.push_back(1);
    yaz();
    return 0;
  }
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= a; j++)
      if (i != j and prime(arr[i] + arr[j])) {
        v.push_back(arr[i]);
        v.push_back(arr[j]);
        yaz();
        return 0;
      }
  v.push_back(arr[1]);
  yaz();
  return 0;
}

### Prompt

In CPP, your task is to solve 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;
template <class T>
bool umin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool umax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
vector<int> v;
int arr[100009], p[100009 << 5];
void yaz() {
  printf("%d\n", int(v.size()));
  for (int i = 0; i < v.size(); i++) printf("%d ", v[i]);
  printf("\n");
}
int prime(int x) {
  p[1] = 2;
  return (p[x] == 1);
}
int main() {
  for (int i = 2; i <= 2000000; i++) {
    if (p[i] == 2) continue;
    p[i] = 1;
    for (int j = i * 2; j <= 2000000; j += i) p[j] = 2;
  }
  int a, bir = 0;
  scanf("%d", &a);
  for (int i = 1; i <= a; i++) {
    scanf("%d", arr + i);
    if (arr[i] == 1) bir++;
  }
  if (bir > 0) {
    for (int i = 1; i <= a; i++)
      if (arr[i] != 1 and prime(arr[i] + 1)) {
        for (int j = 1; j <= bir; j++) v.push_back(1);
        v.push_back(arr[i]);
        yaz();
        return 0;
      }
  }
  if (bir >= 2) {
    for (int i = 1; i <= bir; i++) v.push_back(1);
    yaz();
    return 0;
  }
  for (int i = 1; i <= a; i++)
    for (int j = 1; j <= a; j++)
      if (i != j and prime(arr[i] + arr[j])) {
        v.push_back(arr[i]);
        v.push_back(arr[j]);
        yaz();
        return 0;
      }
  v.push_back(arr[1]);
  yaz();
  return 0;
}

```