#include <bits/stdc++.h>
using namespace std;
bool prime[2000006];
int ara[1003];
void sieve() {
  for (int i = 2; i <= 2000000; i++)
    if (!prime[i]) {
      for (int j = 2 * i; j <= 2000000; j += i) prime[j] = 1;
    }
  return;
}
int main() {
  sieve();
  int n, o = 0, t = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &ara[i]), o += (ara[i] == 1);
    if (ara[i] != 1 && !prime[ara[i] + 1]) t = ara[i];
  }
  if (o && t) {
    printf("%d\n", o + 1);
    for (int i = 0; i < o; i++) printf("1 ");
    printf("%d\n", t);
  } else if (o > 1) {
    printf("%d\n", o);
    for (int i = 0; i < o; i++) printf("1 ");
    puts("");
  } else {
    int a = -1, b = -1;
    for (int i = 0; i < n - 1 && a == -1; i++)
      for (int j = i + 1; j < n && a == -1; j++)
        if (!prime[ara[i] + ara[j]]) a = ara[i], b = ara[j];
    if (a == -1)
      printf("1\n%d\n", ara[0]);
    else
      printf("2\n%d %d\n", a, b);
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
bool prime[2000006];
int ara[1003];
void sieve() {
  for (int i = 2; i <= 2000000; i++)
    if (!prime[i]) {
      for (int j = 2 * i; j <= 2000000; j += i) prime[j] = 1;
    }
  return;
}
int main() {
  sieve();
  int n, o = 0, t = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &ara[i]), o += (ara[i] == 1);
    if (ara[i] != 1 && !prime[ara[i] + 1]) t = ara[i];
  }
  if (o && t) {
    printf("%d\n", o + 1);
    for (int i = 0; i < o; i++) printf("1 ");
    printf("%d\n", t);
  } else if (o > 1) {
    printf("%d\n", o);
    for (int i = 0; i < o; i++) printf("1 ");
    puts("");
  } else {
    int a = -1, b = -1;
    for (int i = 0; i < n - 1 && a == -1; i++)
      for (int j = i + 1; j < n && a == -1; j++)
        if (!prime[ara[i] + ara[j]]) a = ara[i], b = ara[j];
    if (a == -1)
      printf("1\n%d\n", ara[0]);
    else
      printf("2\n%d %d\n", a, b);
  }
  return 0;
}

```