#include <bits/stdc++.h>
using namespace std;
const int mx = 1e7 + 10;
const double eps = 1e-9;
const int inf = 1e9;
int a[mx];
bool prime[mx];
void seive() {
  for (int i = 2; i < mx; i++) prime[i] = 1;
  prime[0] = prime[1] = 0;
  for (int i = 2; i * i < mx; i++)
    if (prime[i])
      for (int j = i * i; j < mx; j += i) prime[j] = 0;
}
vector<int> v;
int main() {
  seive();
  int n;
  scanf("%d", &n);
  int cnt = 0, pos = -1;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    if (a[i] == 1) cnt++;
  }
  bool ok = 0;
  int tmp = 0, tmp2 = 0, ans = -1;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (prime[a[i] + a[j]]) {
        if (a[i] == 1 && a[j] == 1) {
          if (cnt > ans) {
            ans = cnt;
            ok = 1;
          }
        } else if (a[i] == 1 || a[j] == 1) {
          if (cnt + 1 > ans) {
            ans = cnt + 1;
            ok = 1;
            tmp = max(a[i], a[j]);
          }
        } else {
          if (ans < 2) {
            tmp = a[i];
            tmp2 = a[j];
            ans = 2;
          }
        }
      }
    }
  }
  if (ans == -1) {
    printf("1\n%d", a[0]);
    return 0;
  }
  printf("%d\n", ans);
  if (ok) {
    while (cnt--) printf("1 ");
    if (tmp) printf("%d", tmp);
    return 0;
  }
  if (tmp && tmp2) {
    printf("%d %d", tmp, tmp2);
  }
}

### 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 mx = 1e7 + 10;
const double eps = 1e-9;
const int inf = 1e9;
int a[mx];
bool prime[mx];
void seive() {
  for (int i = 2; i < mx; i++) prime[i] = 1;
  prime[0] = prime[1] = 0;
  for (int i = 2; i * i < mx; i++)
    if (prime[i])
      for (int j = i * i; j < mx; j += i) prime[j] = 0;
}
vector<int> v;
int main() {
  seive();
  int n;
  scanf("%d", &n);
  int cnt = 0, pos = -1;
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    if (a[i] == 1) cnt++;
  }
  bool ok = 0;
  int tmp = 0, tmp2 = 0, ans = -1;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (prime[a[i] + a[j]]) {
        if (a[i] == 1 && a[j] == 1) {
          if (cnt > ans) {
            ans = cnt;
            ok = 1;
          }
        } else if (a[i] == 1 || a[j] == 1) {
          if (cnt + 1 > ans) {
            ans = cnt + 1;
            ok = 1;
            tmp = max(a[i], a[j]);
          }
        } else {
          if (ans < 2) {
            tmp = a[i];
            tmp2 = a[j];
            ans = 2;
          }
        }
      }
    }
  }
  if (ans == -1) {
    printf("1\n%d", a[0]);
    return 0;
  }
  printf("%d\n", ans);
  if (ok) {
    while (cnt--) printf("1 ");
    if (tmp) printf("%d", tmp);
    return 0;
  }
  if (tmp && tmp2) {
    printf("%d %d", tmp, tmp2);
  }
}

```