#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 5;
int vis[N], pri[N], pn = 0;
void init() {
  for (int i = 2; i < N; i++) {
    if (!vis[i]) {
      for (int j = i + i; j < N; j += i) vis[j] = 1;
    }
  }
}
int cnt[N], a[N];
int main() {
  init();
  int n;
  while (cin >> n) {
    memset(cnt, 0, sizeof(cnt));
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), cnt[a[i]]++;
    int p = a[1], q, ans = 1;
    for (int i = 1; i <= n; i++) {
      if (cnt[1] && vis[a[i] + 1] == 0 && ans < cnt[1] + 1) {
        if (a[i] != 1)
          ans = cnt[1] + 1;
        else
          ans = cnt[1];
        p = 1, q = a[i];
      }
      for (int j = i + 1; j <= n; j++) {
        if (vis[a[i] + a[j]] == 0) {
          if (ans <= 2) ans = 2, p = a[i], q = a[j];
        }
      }
    }
    cout << ans << endl;
    if (ans > 2) {
      for (int i = 0; i < cnt[1]; i++) printf("1 ");
      if (q != 1) printf("%d", q);
    } else if (ans == 2)
      printf("%d %d", p, q);
    else
      printf("%d", p);
    printf("\n");
  }
  return 0;
}

### Prompt

Your challenge is to write 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 N = 2e6 + 5;
int vis[N], pri[N], pn = 0;
void init() {
  for (int i = 2; i < N; i++) {
    if (!vis[i]) {
      for (int j = i + i; j < N; j += i) vis[j] = 1;
    }
  }
}
int cnt[N], a[N];
int main() {
  init();
  int n;
  while (cin >> n) {
    memset(cnt, 0, sizeof(cnt));
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), cnt[a[i]]++;
    int p = a[1], q, ans = 1;
    for (int i = 1; i <= n; i++) {
      if (cnt[1] && vis[a[i] + 1] == 0 && ans < cnt[1] + 1) {
        if (a[i] != 1)
          ans = cnt[1] + 1;
        else
          ans = cnt[1];
        p = 1, q = a[i];
      }
      for (int j = i + 1; j <= n; j++) {
        if (vis[a[i] + a[j]] == 0) {
          if (ans <= 2) ans = 2, p = a[i], q = a[j];
        }
      }
    }
    cout << ans << endl;
    if (ans > 2) {
      for (int i = 0; i < cnt[1]; i++) printf("1 ");
      if (q != 1) printf("%d", q);
    } else if (ans == 2)
      printf("%d %d", p, q);
    else
      printf("%d", p);
    printf("\n");
  }
  return 0;
}

```