#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 2e6 + 6;
int n, a[1005], one, flag;
bool prime[MAXN];
int main(int argc, char const *argv[]) {
  prime[0] = prime[1] = true;
  for (int i = 2; i < MAXN; ++i)
    if (!prime[i])
      for (int j = i * 2; j < MAXN; j += i) prime[j] = true;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  for (int i = 0; i < n; ++i)
    if (a[i] == 1)
      one++;
    else if (!prime[a[i] + 1])
      flag = a[i];
  if (one + (flag > 0) >= 2) {
    printf("%d\n", one + (flag > 0));
    for (int i = 0; i < one; ++i) printf("%d ", 1);
    if (flag) printf("%d\n", flag);
    return 0;
  }
  for (int i = 0; i < n; ++i)
    if (a[i] & 1)
      for (int j = 0; j < n; ++j)
        if (a[j] % 2 == 0)
          if (!prime[a[i] + a[j]]) return 0 * printf("2\n%d %d\n", a[i], a[j]);
  printf("1\n%d\n", a[0]);
  return 0;
}

### Prompt

Generate 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 MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 2e6 + 6;
int n, a[1005], one, flag;
bool prime[MAXN];
int main(int argc, char const *argv[]) {
  prime[0] = prime[1] = true;
  for (int i = 2; i < MAXN; ++i)
    if (!prime[i])
      for (int j = i * 2; j < MAXN; j += i) prime[j] = true;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
  for (int i = 0; i < n; ++i)
    if (a[i] == 1)
      one++;
    else if (!prime[a[i] + 1])
      flag = a[i];
  if (one + (flag > 0) >= 2) {
    printf("%d\n", one + (flag > 0));
    for (int i = 0; i < one; ++i) printf("%d ", 1);
    if (flag) printf("%d\n", flag);
    return 0;
  }
  for (int i = 0; i < n; ++i)
    if (a[i] & 1)
      for (int j = 0; j < n; ++j)
        if (a[j] % 2 == 0)
          if (!prime[a[i] + a[j]]) return 0 * printf("2\n%d %d\n", a[i], a[j]);
  printf("1\n%d\n", a[0]);
  return 0;
}

```