#include <bits/stdc++.h>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {1, -1, 1, 0, -1, 0, 1, -1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
long long int mark[2000005], a[2000005];
void sieve() {
  long long int i, j;
  mark[1] = 1;
  for (i = 4; i < 2000005; i += 2) mark[i] = 1;
  for (i = 3; i < 2000005; i += 2) {
    if (mark[i] == 0) {
      for (j = i + i; j < 2000005; j += i) mark[j] = 1;
    }
  }
}
int main() {
  long long int n, i, j;
  sieve();
  while (cin >> n) {
    long long int f = 0, cn = 0;
    long long int ex = -1;
    for (long long int i = 1; i < n + 1; i++) {
      cin >> a[i];
      if (a[i] == 1) cn++;
      if (mark[a[i] + 1] == 0 && a[i] != 1) ex = a[i];
    }
    if (cn > 0 && ex != -1) {
      cout << cn + 1 << endl;
      for (long long int i = 1; i < cn + 1; i++) cout << "1 ";
      cout << ex << endl;
      continue;
    }
    if (cn > 1) {
      cout << cn << endl;
      for (long long int i = 1; i < cn + 1; i++) cout << "1 ";
      cout << endl;
      return 0;
    }
    for (long long int i = 1; i < n + 1; i++) {
      for (long long int j = i + 1; j < n + 1; j++) {
        long long int p = a[i] + a[j];
        if (mark[p] == 0) {
          cout << "2" << endl;
          cout << a[i] << " " << a[j] << endl;
          return 0;
        }
      }
    }
    cout << "1" << endl;
    cout << a[1] << endl;
  }
  return 0;
}

### Prompt

Please formulate 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>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {1, -1, 1, 0, -1, 0, 1, -1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
long long int mark[2000005], a[2000005];
void sieve() {
  long long int i, j;
  mark[1] = 1;
  for (i = 4; i < 2000005; i += 2) mark[i] = 1;
  for (i = 3; i < 2000005; i += 2) {
    if (mark[i] == 0) {
      for (j = i + i; j < 2000005; j += i) mark[j] = 1;
    }
  }
}
int main() {
  long long int n, i, j;
  sieve();
  while (cin >> n) {
    long long int f = 0, cn = 0;
    long long int ex = -1;
    for (long long int i = 1; i < n + 1; i++) {
      cin >> a[i];
      if (a[i] == 1) cn++;
      if (mark[a[i] + 1] == 0 && a[i] != 1) ex = a[i];
    }
    if (cn > 0 && ex != -1) {
      cout << cn + 1 << endl;
      for (long long int i = 1; i < cn + 1; i++) cout << "1 ";
      cout << ex << endl;
      continue;
    }
    if (cn > 1) {
      cout << cn << endl;
      for (long long int i = 1; i < cn + 1; i++) cout << "1 ";
      cout << endl;
      return 0;
    }
    for (long long int i = 1; i < n + 1; i++) {
      for (long long int j = i + 1; j < n + 1; j++) {
        long long int p = a[i] + a[j];
        if (mark[p] == 0) {
          cout << "2" << endl;
          cout << a[i] << " " << a[j] << endl;
          return 0;
        }
      }
    }
    cout << "1" << endl;
    cout << a[1] << endl;
  }
  return 0;
}

```