#include <bits/stdc++.h>
using namespace std;
const int N = 1005;
const int M = 2000005;
bitset<M + 5> isprime;
void getPrime() {
  isprime.set();
  isprime[0] = isprime[1] = false;
  for (int i = 2; i <= M; i++) {
    if (isprime[i]) {
      for (long long j = (long long)i * i; j <= M; j += i) {
        isprime[j] = false;
      }
    }
  }
}
int n;
int a[N], b[N];
int main() {
  getPrime();
  while (~scanf("%d", &n)) {
    int ans = 0, one = 0, other = 0;
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      for (int j = 0; !ans && j < i; j++) {
        if (isprime[a[i] + a[j]]) {
          ans = 2;
          b[0] = a[i];
          b[1] = a[j];
        }
      }
      if (a[i] == 1) {
        one++;
      }
      if (a[i] != 1 && isprime[a[i] + 1]) {
        other = a[i];
      }
    }
    if (one + (other != 0) > ans) {
      ans = one + (other != 0);
      fill(b, b + one, 1);
      if (other != 0) {
        b[ans - 1] = other;
      }
    }
    if (ans == 0) {
      ans++;
      b[0] = a[0];
    }
    printf("%d\n", ans);
    for (int i = 0; i < ans; i++) {
      printf("%d%c", b[i], i == ans - 1 ? '\n' : ' ');
    }
  }
}

### 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 N = 1005;
const int M = 2000005;
bitset<M + 5> isprime;
void getPrime() {
  isprime.set();
  isprime[0] = isprime[1] = false;
  for (int i = 2; i <= M; i++) {
    if (isprime[i]) {
      for (long long j = (long long)i * i; j <= M; j += i) {
        isprime[j] = false;
      }
    }
  }
}
int n;
int a[N], b[N];
int main() {
  getPrime();
  while (~scanf("%d", &n)) {
    int ans = 0, one = 0, other = 0;
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      for (int j = 0; !ans && j < i; j++) {
        if (isprime[a[i] + a[j]]) {
          ans = 2;
          b[0] = a[i];
          b[1] = a[j];
        }
      }
      if (a[i] == 1) {
        one++;
      }
      if (a[i] != 1 && isprime[a[i] + 1]) {
        other = a[i];
      }
    }
    if (one + (other != 0) > ans) {
      ans = one + (other != 0);
      fill(b, b + one, 1);
      if (other != 0) {
        b[ans - 1] = other;
      }
    }
    if (ans == 0) {
      ans++;
      b[0] = a[0];
    }
    printf("%d\n", ans);
    for (int i = 0; i < ans; i++) {
      printf("%d%c", b[i], i == ans - 1 ? '\n' : ' ');
    }
  }
}

```