#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2 * (1e6 + 10), BASE = 3137, MOD = 1e9 + 7, inf = 1000000009;
const double PI = 3.1415926535;
const int dx[] = {0, -1, 0, 1, 1, -1, 1, -1};
const int dy[] = {1, 0, -1, 0, 1, -1, -1, 1};
int n;
int arr[1010];
int prosti[MAXN];
int sol;
vector<int> ans;
void Eratosten() {
  for (int i = 0; i < MAXN; i++) prosti[i] = 1;
  for (int i = 2; i < MAXN; i++)
    if (prosti[i] == 1)
      for (int j = i + i; j < MAXN; j += i) prosti[j] = 0;
}
int main() {
  Eratosten();
  scanf("%d", &n);
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    if (arr[i] == 1) cnt++;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      if (prosti[arr[i] + arr[j]]) {
        sol = 2;
        ans.push_back(arr[i]);
        ans.push_back(arr[j]);
        break;
      }
    }
    if (sol) break;
  }
  vector<int> ans2(cnt, 1);
  int sol2 = cnt;
  int pamti = -1;
  for (int i = 0; i < n; i++) {
    if (prosti[arr[i] + 1] && cnt && arr[i] != 1) {
      sol2++;
      pamti = arr[i];
      break;
    }
  }
  bool flag = false;
  if (sol2 > sol) {
    sol = sol2;
    ans = ans2;
    if (pamti != -1) ans.push_back(pamti);
    printf("%d\n", (int)ans.size());
    for (int i = 0; i < (int)ans.size(); i++) printf("%d ", ans[i]);
    flag = true;
  } else if ((int)ans.size() != 0) {
    printf("%d\n", (int)ans.size());
    for (int i = 0; i < (int)ans.size(); i++) printf("%d ", ans[i]);
    flag = true;
  } else {
    for (int i = 0; i < n; i++) {
      if (prosti[arr[i]]) {
        printf("1\n");
        printf("%d\n", arr[i]);
        flag = true;
        break;
      }
    }
  }
  if (!flag) printf("1\n%d\n", arr[0]);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 MAXN = 2 * (1e6 + 10), BASE = 3137, MOD = 1e9 + 7, inf = 1000000009;
const double PI = 3.1415926535;
const int dx[] = {0, -1, 0, 1, 1, -1, 1, -1};
const int dy[] = {1, 0, -1, 0, 1, -1, -1, 1};
int n;
int arr[1010];
int prosti[MAXN];
int sol;
vector<int> ans;
void Eratosten() {
  for (int i = 0; i < MAXN; i++) prosti[i] = 1;
  for (int i = 2; i < MAXN; i++)
    if (prosti[i] == 1)
      for (int j = i + i; j < MAXN; j += i) prosti[j] = 0;
}
int main() {
  Eratosten();
  scanf("%d", &n);
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    if (arr[i] == 1) cnt++;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      if (prosti[arr[i] + arr[j]]) {
        sol = 2;
        ans.push_back(arr[i]);
        ans.push_back(arr[j]);
        break;
      }
    }
    if (sol) break;
  }
  vector<int> ans2(cnt, 1);
  int sol2 = cnt;
  int pamti = -1;
  for (int i = 0; i < n; i++) {
    if (prosti[arr[i] + 1] && cnt && arr[i] != 1) {
      sol2++;
      pamti = arr[i];
      break;
    }
  }
  bool flag = false;
  if (sol2 > sol) {
    sol = sol2;
    ans = ans2;
    if (pamti != -1) ans.push_back(pamti);
    printf("%d\n", (int)ans.size());
    for (int i = 0; i < (int)ans.size(); i++) printf("%d ", ans[i]);
    flag = true;
  } else if ((int)ans.size() != 0) {
    printf("%d\n", (int)ans.size());
    for (int i = 0; i < (int)ans.size(); i++) printf("%d ", ans[i]);
    flag = true;
  } else {
    for (int i = 0; i < n; i++) {
      if (prosti[arr[i]]) {
        printf("1\n");
        printf("%d\n", arr[i]);
        flag = true;
        break;
      }
    }
  }
  if (!flag) printf("1\n%d\n", arr[0]);
  return 0;
}

```