#include <bits/stdc++.h>
using namespace std;
int a[1010];
int n;
bool is_prime(int x) {
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) return false;
  }
  return true;
}
int main() {
  scanf("%d", &n);
  int num[1010];
  int p = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) num[p++] = 1;
  }
  sort(a, a + n);
  if (p > 1) {
    for (int i = p; i < n; i++) {
      if (is_prime(a[i] + 1)) {
        num[p++] = a[i];
        break;
      }
    }
  } else if (p == 1) {
    for (int i = 1; i < n; i++) {
      if (is_prime(a[i] + 1)) {
        num[p++] = a[i];
        break;
      }
    }
    if (p == 1) {
      int m = 0;
      int b = 0;
      for (int i = 1; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
          if (is_prime(a[i] + a[j])) {
            m = a[i];
            b = a[j];
            break;
          }
        }
      }
      if (m != 0) {
        p = 0;
        num[p++] = m;
        num[p++] = b;
      }
    }
  } else {
    int flag = 0;
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        if (is_prime(a[i] + a[j])) {
          num[p++] = a[i];
          num[p++] = a[j];
          flag = 1;
          break;
        }
      }
      if (flag) break;
    }
    if (p == 0) {
      num[p++] = a[0];
    }
  }
  printf("%d\n", p);
  for (int i = 0; i < p; i++) {
    if (i == p - 1)
      printf("%d\n", num[i]);
    else
      printf("%d ", num[i]);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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[1010];
int n;
bool is_prime(int x) {
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) return false;
  }
  return true;
}
int main() {
  scanf("%d", &n);
  int num[1010];
  int p = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) num[p++] = 1;
  }
  sort(a, a + n);
  if (p > 1) {
    for (int i = p; i < n; i++) {
      if (is_prime(a[i] + 1)) {
        num[p++] = a[i];
        break;
      }
    }
  } else if (p == 1) {
    for (int i = 1; i < n; i++) {
      if (is_prime(a[i] + 1)) {
        num[p++] = a[i];
        break;
      }
    }
    if (p == 1) {
      int m = 0;
      int b = 0;
      for (int i = 1; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
          if (is_prime(a[i] + a[j])) {
            m = a[i];
            b = a[j];
            break;
          }
        }
      }
      if (m != 0) {
        p = 0;
        num[p++] = m;
        num[p++] = b;
      }
    }
  } else {
    int flag = 0;
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        if (is_prime(a[i] + a[j])) {
          num[p++] = a[i];
          num[p++] = a[j];
          flag = 1;
          break;
        }
      }
      if (flag) break;
    }
    if (p == 0) {
      num[p++] = a[0];
    }
  }
  printf("%d\n", p);
  for (int i = 0; i < p; i++) {
    if (i == p - 1)
      printf("%d\n", num[i]);
    else
      printf("%d ", num[i]);
  }
  return 0;
}

```