#include <bits/stdc++.h>
using namespace std;
vector<int> p;
int is_p[2000010];
int a[2000010];
int main() {
  int i, j;
  for (i = 2; i < 2000010; i++) is_p[i] = 1;
  for (i = 2; i <= 2000010; i++) {
    if (is_p[i]) p.push_back(i);
    for (j = 0; j < p.size(); j++) {
      if (i * p[j] >= 2000010) break;
      is_p[i * p[j]] = 0;
      if (i % p[j] == 0) break;
    }
  }
  int n, cnt = 0;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) cnt++;
  }
  int ans = 0;
  if (cnt <= 1) {
    for (i = 1; i <= n; i++) {
      for (j = i + 1; j <= n; j++) {
        if (is_p[a[i] + a[j]]) {
          printf("2\n%d %d\n", a[i], a[j]);
          return 0;
        }
      }
    }
    printf("1\n%d\n", a[1]);
  } else {
    for (i = 1; i <= n; i++) {
      if (is_p[1 + a[i]] && a[i] != 1) {
        ans = a[i];
        printf("%d\n", cnt + 1);
        for (j = 0; j < cnt; j++) printf("1 ");
        printf("%d\n", ans);
        return 0;
      }
    }
    printf("%d\n", cnt);
    for (j = 0; j < cnt; j++) printf("1 ");
  }
  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;
vector<int> p;
int is_p[2000010];
int a[2000010];
int main() {
  int i, j;
  for (i = 2; i < 2000010; i++) is_p[i] = 1;
  for (i = 2; i <= 2000010; i++) {
    if (is_p[i]) p.push_back(i);
    for (j = 0; j < p.size(); j++) {
      if (i * p[j] >= 2000010) break;
      is_p[i * p[j]] = 0;
      if (i % p[j] == 0) break;
    }
  }
  int n, cnt = 0;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) cnt++;
  }
  int ans = 0;
  if (cnt <= 1) {
    for (i = 1; i <= n; i++) {
      for (j = i + 1; j <= n; j++) {
        if (is_p[a[i] + a[j]]) {
          printf("2\n%d %d\n", a[i], a[j]);
          return 0;
        }
      }
    }
    printf("1\n%d\n", a[1]);
  } else {
    for (i = 1; i <= n; i++) {
      if (is_p[1 + a[i]] && a[i] != 1) {
        ans = a[i];
        printf("%d\n", cnt + 1);
        for (j = 0; j < cnt; j++) printf("1 ");
        printf("%d\n", ans);
        return 0;
      }
    }
    printf("%d\n", cnt);
    for (j = 0; j < cnt; j++) printf("1 ");
  }
  return 0;
}

```