#include <bits/stdc++.h>
using namespace std;
int n, v[1005], s1, s2, s3, x[2], y[1005], z;
bool yes[2000001], a;
int main() {
  memset(yes, 1, sizeof yes);
  for (int i = 2; i <= 1500; i++) {
    if (yes[i]) {
      for (int j = i * i; j <= 2000000; j += i) yes[j] = false;
    }
  }
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &v[i]);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++)
      if ((i != j) && (yes[v[i] + v[j]])) {
        s1 = 2;
        x[0] = v[i], x[1] = v[j];
        break;
      }
  }
  for (int i = 0; i < n; i++) {
    if (v[i] == 1)
      y[s2++] = v[i];
    else if (!a && yes[v[i] + 1])
      a = true, y[s2++] = v[i];
  }
  if (s1 == s2 && s1 == 0)
    printf("1\n%d\n", v[0]);
  else if (s1 > s2)
    printf("%d\n%d %d\n", s1, x[0], x[1]);
  else {
    printf("%d\n", s2);
    for (int i = 0; i < s2; i++) {
      if (i != 0) printf(" ");
      printf("%d", y[i]);
    }
    printf("\n");
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 n, v[1005], s1, s2, s3, x[2], y[1005], z;
bool yes[2000001], a;
int main() {
  memset(yes, 1, sizeof yes);
  for (int i = 2; i <= 1500; i++) {
    if (yes[i]) {
      for (int j = i * i; j <= 2000000; j += i) yes[j] = false;
    }
  }
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &v[i]);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++)
      if ((i != j) && (yes[v[i] + v[j]])) {
        s1 = 2;
        x[0] = v[i], x[1] = v[j];
        break;
      }
  }
  for (int i = 0; i < n; i++) {
    if (v[i] == 1)
      y[s2++] = v[i];
    else if (!a && yes[v[i] + 1])
      a = true, y[s2++] = v[i];
  }
  if (s1 == s2 && s1 == 0)
    printf("1\n%d\n", v[0]);
  else if (s1 > s2)
    printf("%d\n%d %d\n", s1, x[0], x[1]);
  else {
    printf("%d\n", s2);
    for (int i = 0; i < s2; i++) {
      if (i != 0) printf(" ");
      printf("%d", y[i]);
    }
    printf("\n");
  }
  return 0;
}

```