#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e6 + 5;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
bool vis[maxn];
int a[1005], n, cnt;
struct edge {
  int v, next;
} e[maxn];
void init() {
  vis[1] = 1;
  for (int i = 2; i < maxn; i++)
    if (!vis[i]) {
      for (int k = i + i; k < maxn; k += i) vis[k] = 1;
    }
}
int main() {
  init();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) cnt++;
  }
  if (cnt == 0) {
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j < i; j++) {
        if (!vis[a[i] + a[j]]) {
          printf("%d\n%d %d\n", 2, a[i], a[j]);
          return 0;
        }
      }
    }
    printf("%d\n%d\n", 1, a[1]);
    return 0;
  } else if (cnt == 1) {
    for (int i = 1; i <= n; i++) {
      if (a[i] == 1) continue;
      for (int j = 1; j < i; j++) {
        if (a[j] == 1) continue;
        if (!vis[a[i] + a[j]]) {
          printf("%d\n%d %d\n", 2, a[i], a[j]);
          return 0;
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1) continue;
    if (!vis[a[i] + 1]) {
      printf("%d\n%d", 1 + cnt, a[i]);
      while (cnt--) printf(" 1");
      puts("");
      return 0;
    }
  }
  printf("%d\n", cnt);
  while (cnt--) printf("1 ");
  puts("");
  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 maxn = 2e6 + 5;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
bool vis[maxn];
int a[1005], n, cnt;
struct edge {
  int v, next;
} e[maxn];
void init() {
  vis[1] = 1;
  for (int i = 2; i < maxn; i++)
    if (!vis[i]) {
      for (int k = i + i; k < maxn; k += i) vis[k] = 1;
    }
}
int main() {
  init();
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) cnt++;
  }
  if (cnt == 0) {
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j < i; j++) {
        if (!vis[a[i] + a[j]]) {
          printf("%d\n%d %d\n", 2, a[i], a[j]);
          return 0;
        }
      }
    }
    printf("%d\n%d\n", 1, a[1]);
    return 0;
  } else if (cnt == 1) {
    for (int i = 1; i <= n; i++) {
      if (a[i] == 1) continue;
      for (int j = 1; j < i; j++) {
        if (a[j] == 1) continue;
        if (!vis[a[i] + a[j]]) {
          printf("%d\n%d %d\n", 2, a[i], a[j]);
          return 0;
        }
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1) continue;
    if (!vis[a[i] + 1]) {
      printf("%d\n%d", 1 + cnt, a[i]);
      while (cnt--) printf(" 1");
      puts("");
      return 0;
    }
  }
  printf("%d\n", cnt);
  while (cnt--) printf("1 ");
  puts("");
  return 0;
}

```