#include <bits/stdc++.h>
using namespace std;
long long int powmod(long long int a, long long int b) {
  if (b == 0) return 1;
  long long int x = powmod(a, b / 2);
  long long int y = (x * x) % 1000000007;
  if (b % 2) return (a * y) % 1000000007;
  return y % 1000000007;
}
bool isprime[2000010];
int a[1010];
void sieve() {
  for (int i = 0; i < 2000010; i++) isprime[i] = true;
  isprime[0] = isprime[1] = false;
  for (long long int i = 2; i < 2000010; i++) {
    if (isprime[i]) {
      for (long long int j = 2; j * i < 2000010; j++) isprime[i * j] = false;
    }
  }
}
vector<int> ans;
int main() {
  sieve();
  int n;
  scanf("%d", &n);
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) {
      ans.push_back(1);
      cnt++;
    }
  }
  int ind1 = -1;
  int ind2 = -1;
  bool flag = false;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if ((a[j] % 2 != 0 && a[i] % 2 == 0)) {
        if (isprime[a[i] + a[j]]) {
          if (a[j] == 1) {
            ans.push_back(a[i]);
            flag = true;
            break;
          }
          ind1 = i;
          ind2 = j;
        }
      }
    }
    if (flag) break;
  }
  if (flag) {
    printf("%d\n", (int)(ans.size()));
    for (int i = 0; i < (int)(ans.size()); i++) printf("%d ", ans[i]);
    printf("\n");
    return 0;
  }
  if (cnt >= 2) {
    printf("%d\n", cnt);
    for (int i = 0; i < cnt; i++) printf("1 ");
    printf("\n");
    return 0;
  }
  if (ind1 != -1 && ind2 != -1) {
    printf("2\n");
    printf("%d %d\n", a[ind1], a[ind2]);
    return 0;
  }
  printf("1\n");
  printf("%d\n", a[0]);
  return 0;
}

### Prompt

Please create a solution in CPP 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;
long long int powmod(long long int a, long long int b) {
  if (b == 0) return 1;
  long long int x = powmod(a, b / 2);
  long long int y = (x * x) % 1000000007;
  if (b % 2) return (a * y) % 1000000007;
  return y % 1000000007;
}
bool isprime[2000010];
int a[1010];
void sieve() {
  for (int i = 0; i < 2000010; i++) isprime[i] = true;
  isprime[0] = isprime[1] = false;
  for (long long int i = 2; i < 2000010; i++) {
    if (isprime[i]) {
      for (long long int j = 2; j * i < 2000010; j++) isprime[i * j] = false;
    }
  }
}
vector<int> ans;
int main() {
  sieve();
  int n;
  scanf("%d", &n);
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    if (a[i] == 1) {
      ans.push_back(1);
      cnt++;
    }
  }
  int ind1 = -1;
  int ind2 = -1;
  bool flag = false;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if ((a[j] % 2 != 0 && a[i] % 2 == 0)) {
        if (isprime[a[i] + a[j]]) {
          if (a[j] == 1) {
            ans.push_back(a[i]);
            flag = true;
            break;
          }
          ind1 = i;
          ind2 = j;
        }
      }
    }
    if (flag) break;
  }
  if (flag) {
    printf("%d\n", (int)(ans.size()));
    for (int i = 0; i < (int)(ans.size()); i++) printf("%d ", ans[i]);
    printf("\n");
    return 0;
  }
  if (cnt >= 2) {
    printf("%d\n", cnt);
    for (int i = 0; i < cnt; i++) printf("1 ");
    printf("\n");
    return 0;
  }
  if (ind1 != -1 && ind2 != -1) {
    printf("2\n");
    printf("%d %d\n", a[ind1], a[ind2]);
    return 0;
  }
  printf("1\n");
  printf("%d\n", a[0]);
  return 0;
}

```