#include <bits/stdc++.h>
using namespace std;
int dx[] = {0, 0, -1, 1, -1, -1, 1, 1};
int dy[] = {1, -1, 0, 0, -1, 1, -1, 1};
int year[15] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
bool prime[10000005];
long long M = 1e9 + 7;
long long fast_power(long long b, long long p) {
  long long res = 1;
  while (p > 0) {
    if (p & 1) res = (res * b);
    p /= 2, b = (b * b);
  }
  return res;
}
bool p[2000005];
int arr[1000005];
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  for (int i = 2; i < 2000005; ++i) p[i] = 1;
  for (int i = 2; i * i < 2000005; ++i) {
    if (p[i]) {
      for (int j = i + i; j < 2000005; j += i) p[j] = 0;
    }
  }
  int n, x;
  cin >> n;
  vector<int> v;
  while (n--) {
    cin >> x;
    if (!arr[x]) v.push_back(x);
    ++arr[x];
  }
  n = v.size();
  sort(v.begin(), v.end());
  bool f = 0;
  int a = 0, b = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      if (p[(v[i] + v[j])]) {
        f = 1, a = i, b = j;
        break;
      }
    }
    if (f) break;
  }
  if (f) {
    if (v[a] != 1) {
      if (v[0] == 1 && arr[1] > 1) {
        cout << arr[1] << "\n";
        for (int i = 0; i < arr[1]; ++i) cout << "1 ";
      } else {
        cout << 2 << "\n";
        cout << v[a] << " " << v[b];
      }
    } else {
      cout << arr[1] + 1 << "\n";
      for (int i = 0; i < arr[1]; ++i) cout << "1 ";
      cout << v[b];
    }
  } else {
    if (v[0] == 1) {
      cout << arr[1] << "\n";
      for (int i = 0; i < arr[1]; ++i) cout << "1 ";
    } else
      cout << "1\n" << v[0];
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
int dx[] = {0, 0, -1, 1, -1, -1, 1, 1};
int dy[] = {1, -1, 0, 0, -1, 1, -1, 1};
int year[15] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
bool prime[10000005];
long long M = 1e9 + 7;
long long fast_power(long long b, long long p) {
  long long res = 1;
  while (p > 0) {
    if (p & 1) res = (res * b);
    p /= 2, b = (b * b);
  }
  return res;
}
bool p[2000005];
int arr[1000005];
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  for (int i = 2; i < 2000005; ++i) p[i] = 1;
  for (int i = 2; i * i < 2000005; ++i) {
    if (p[i]) {
      for (int j = i + i; j < 2000005; j += i) p[j] = 0;
    }
  }
  int n, x;
  cin >> n;
  vector<int> v;
  while (n--) {
    cin >> x;
    if (!arr[x]) v.push_back(x);
    ++arr[x];
  }
  n = v.size();
  sort(v.begin(), v.end());
  bool f = 0;
  int a = 0, b = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      if (p[(v[i] + v[j])]) {
        f = 1, a = i, b = j;
        break;
      }
    }
    if (f) break;
  }
  if (f) {
    if (v[a] != 1) {
      if (v[0] == 1 && arr[1] > 1) {
        cout << arr[1] << "\n";
        for (int i = 0; i < arr[1]; ++i) cout << "1 ";
      } else {
        cout << 2 << "\n";
        cout << v[a] << " " << v[b];
      }
    } else {
      cout << arr[1] + 1 << "\n";
      for (int i = 0; i < arr[1]; ++i) cout << "1 ";
      cout << v[b];
    }
  } else {
    if (v[0] == 1) {
      cout << arr[1] << "\n";
      for (int i = 0; i < arr[1]; ++i) cout << "1 ";
    } else
      cout << "1\n" << v[0];
  }
  return 0;
}

```