#include <bits/stdc++.h>
using namespace std;
int a[200005];
vector<int> ans;
int n;
bool ok(int x) {
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) return false;
  }
  return true;
}
bool solven() {
  ans.clear();
  int flg = 0;
  int count = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1) {
      count++;
      ans.push_back(a[i]);
      continue;
    }
    if (!flg && ok(a[i] + 1)) {
      flg = 1;
      ans.push_back(a[i]);
    }
  }
  if (count == 0) return 0;
  if (ans.size() > 2) return 1;
  return 0;
}
bool solve2() {
  ans.clear();
  int flg = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (ok(a[i] + a[j])) {
        flg = 1;
        ans.push_back(a[i]);
        ans.push_back(a[j]);
        break;
      }
    }
    if (flg) break;
  }
  if (ans.size() == 2) return 1;
  return 0;
}
bool solve1() {
  ans.clear();
  for (int i = 1; i <= n; i++) {
    if (ok(a[i])) {
      ans.push_back(a[i]);
      break;
    }
  }
  if (ans.size() == 1) return 1;
  return 0;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  if (solven()) {
    printf("%d\n", ans.size());
    for (int i = 0; i < ans.size(); i++) {
      printf("%d ", ans[i]);
    }
    printf("\n");
  } else if (solve2()) {
    printf("2\n");
    for (int i = 0; i < ans.size(); i++) {
      printf("%d ", ans[i]);
    }
    printf("\n");
  } else if (solve1()) {
    printf("1\n");
    for (int i = 0; i < ans.size(); i++) {
      printf("%d ", ans[i]);
    }
    printf("\n");
  } else {
    printf("1\n%d\n", a[1]);
  }
}

### 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;
int a[200005];
vector<int> ans;
int n;
bool ok(int x) {
  for (int i = 2; i * i <= x; i++) {
    if (x % i == 0) return false;
  }
  return true;
}
bool solven() {
  ans.clear();
  int flg = 0;
  int count = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1) {
      count++;
      ans.push_back(a[i]);
      continue;
    }
    if (!flg && ok(a[i] + 1)) {
      flg = 1;
      ans.push_back(a[i]);
    }
  }
  if (count == 0) return 0;
  if (ans.size() > 2) return 1;
  return 0;
}
bool solve2() {
  ans.clear();
  int flg = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (ok(a[i] + a[j])) {
        flg = 1;
        ans.push_back(a[i]);
        ans.push_back(a[j]);
        break;
      }
    }
    if (flg) break;
  }
  if (ans.size() == 2) return 1;
  return 0;
}
bool solve1() {
  ans.clear();
  for (int i = 1; i <= n; i++) {
    if (ok(a[i])) {
      ans.push_back(a[i]);
      break;
    }
  }
  if (ans.size() == 1) return 1;
  return 0;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  if (solven()) {
    printf("%d\n", ans.size());
    for (int i = 0; i < ans.size(); i++) {
      printf("%d ", ans[i]);
    }
    printf("\n");
  } else if (solve2()) {
    printf("2\n");
    for (int i = 0; i < ans.size(); i++) {
      printf("%d ", ans[i]);
    }
    printf("\n");
  } else if (solve1()) {
    printf("1\n");
    for (int i = 0; i < ans.size(); i++) {
      printf("%d ", ans[i]);
    }
    printf("\n");
  } else {
    printf("1\n%d\n", a[1]);
  }
}

```