#include <bits/stdc++.h>
using namespace std;
const int mx = 2e6 + 5;
int a[1005];
int pri[mx], vs[mx], pn;
void prim() {
  vs[0] = vs[1] = 1;
  pn = 0;
  for (int i = 2; i < 2000000; i++) {
    if (!vs[i]) pri[pn++] = i;
    for (int j = 0; j < pn; j++) {
      if (i * pri[j] > 2000000) break;
      vs[i * pri[j]] = 1;
      if (i % pri[j] == 0) break;
    }
  }
}
int main() {
  prim();
  int n;
  scanf("%d", &n);
  int cut = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) cut++;
  }
  bool flag = false;
  int ans1, ans2;
  if (cut > 1) {
    for (int i = 0; i < n; i++) {
      if (a[i] == 1) continue;
      if (!vs[a[i] + 1]) {
        ans1 = a[i];
        flag = true;
        break;
      }
    }
    if (flag) {
      printf("%d\n", cut + 1);
      for (int i = 0; i < cut; i++) printf("1 ");
      printf("%d\n", ans1);
    } else {
      printf("%d\n", cut);
      for (int i = 0; i < cut; i++) {
        if (i == cut - 1)
          printf("1\n");
        else
          printf("1 ");
      }
    }
  } else {
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        if (!vs[a[i] + a[j]]) {
          ans1 = a[i];
          ans2 = a[j];
          flag = true;
          break;
        }
      }
    }
    if (flag)
      printf("2\n%d %d\n", ans1, ans2);
    else
      printf("1\n%d\n", a[0]);
  }
}

### Prompt

Your task is to create 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 mx = 2e6 + 5;
int a[1005];
int pri[mx], vs[mx], pn;
void prim() {
  vs[0] = vs[1] = 1;
  pn = 0;
  for (int i = 2; i < 2000000; i++) {
    if (!vs[i]) pri[pn++] = i;
    for (int j = 0; j < pn; j++) {
      if (i * pri[j] > 2000000) break;
      vs[i * pri[j]] = 1;
      if (i % pri[j] == 0) break;
    }
  }
}
int main() {
  prim();
  int n;
  scanf("%d", &n);
  int cut = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) cut++;
  }
  bool flag = false;
  int ans1, ans2;
  if (cut > 1) {
    for (int i = 0; i < n; i++) {
      if (a[i] == 1) continue;
      if (!vs[a[i] + 1]) {
        ans1 = a[i];
        flag = true;
        break;
      }
    }
    if (flag) {
      printf("%d\n", cut + 1);
      for (int i = 0; i < cut; i++) printf("1 ");
      printf("%d\n", ans1);
    } else {
      printf("%d\n", cut);
      for (int i = 0; i < cut; i++) {
        if (i == cut - 1)
          printf("1\n");
        else
          printf("1 ");
      }
    }
  } else {
    for (int i = 0; i < n; i++) {
      for (int j = i + 1; j < n; j++) {
        if (!vs[a[i] + a[j]]) {
          ans1 = a[i];
          ans2 = a[j];
          flag = true;
          break;
        }
      }
    }
    if (flag)
      printf("2\n%d %d\n", ans1, ans2);
    else
      printf("1\n%d\n", a[0]);
  }
}

```