#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("Ofast")
#pragma GCC optimization("unroll-loops")
using namespace std;
long long pow2(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = ((res % 998244353) * (a % 998244353)) % 998244353;
    a = ((a % 998244353) * (a % 998244353));
    b >>= 1;
  }
  return res;
}
map<long long, long long> Linearsieve(long long n) {
  vector<bool> v(n + 1, true);
  vector<long long> primes;
  map<long long, long long> temp;
  for (long long i = 2; i <= n; i++) {
    if (v[i]) {
      primes.push_back(i);
      temp[i]++;
    }
    for (int j = 0; j < primes.size() && i * primes[j] <= n; j++) {
      v[i * primes[j]] = false;
      if (i % primes[j] == 0) {
        break;
      }
    }
  }
  return temp;
}
struct comp {
  bool operator()(pair<long long, long long> a, pair<long long, long long> b) {
    return a.second > b.second;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  map<long long, long long> primes = Linearsieve(2000000);
  int t = 1;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> v(n);
    int f[2] = {0, 0};
    for (long long i = 0; i < (n); ++i) {
      long long x;
      cin >> x;
      v[i] = x;
      if (v[i] == 1) f[1]++;
    }
    int flag = 0;
    int found = 0;
    if (f[1]) {
      for (int i = 0; i < n; i++) {
        if (v[i] != 1 and primes[v[i] + 1]) {
          found = 1;
          flag = i;
          break;
        }
      }
      if (found) {
        cout << f[1] + 1 << endl;
        for (long long i = 0; i < (f[1]); ++i) {
          cout << "1 ";
        }
        cout << v[flag] << endl;
        return 0;
      } else if (f[1] > 1) {
        cout << f[1] << endl;
        for (long long i = 0; i < (f[1]); ++i) {
          cout << "1 ";
        }
        cout << "" << endl;
        return 0;
      }
    }
    for (long long i = 0; i < (n); ++i) {
      for (long long j = (i + 1); j < (n); ++j) {
        if (primes[v[i] + v[j]]) {
          cout << "2\n";
          cout << v[i] << ' ' << v[j] << endl;
          return 0;
        }
      }
    }
    cout << 1 << endl;
    cout << v[0] << endl;
    return 0;
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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>
#pragma GCC target("avx2")
#pragma GCC optimization("Ofast")
#pragma GCC optimization("unroll-loops")
using namespace std;
long long pow2(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = ((res % 998244353) * (a % 998244353)) % 998244353;
    a = ((a % 998244353) * (a % 998244353));
    b >>= 1;
  }
  return res;
}
map<long long, long long> Linearsieve(long long n) {
  vector<bool> v(n + 1, true);
  vector<long long> primes;
  map<long long, long long> temp;
  for (long long i = 2; i <= n; i++) {
    if (v[i]) {
      primes.push_back(i);
      temp[i]++;
    }
    for (int j = 0; j < primes.size() && i * primes[j] <= n; j++) {
      v[i * primes[j]] = false;
      if (i % primes[j] == 0) {
        break;
      }
    }
  }
  return temp;
}
struct comp {
  bool operator()(pair<long long, long long> a, pair<long long, long long> b) {
    return a.second > b.second;
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  map<long long, long long> primes = Linearsieve(2000000);
  int t = 1;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> v(n);
    int f[2] = {0, 0};
    for (long long i = 0; i < (n); ++i) {
      long long x;
      cin >> x;
      v[i] = x;
      if (v[i] == 1) f[1]++;
    }
    int flag = 0;
    int found = 0;
    if (f[1]) {
      for (int i = 0; i < n; i++) {
        if (v[i] != 1 and primes[v[i] + 1]) {
          found = 1;
          flag = i;
          break;
        }
      }
      if (found) {
        cout << f[1] + 1 << endl;
        for (long long i = 0; i < (f[1]); ++i) {
          cout << "1 ";
        }
        cout << v[flag] << endl;
        return 0;
      } else if (f[1] > 1) {
        cout << f[1] << endl;
        for (long long i = 0; i < (f[1]); ++i) {
          cout << "1 ";
        }
        cout << "" << endl;
        return 0;
      }
    }
    for (long long i = 0; i < (n); ++i) {
      for (long long j = (i + 1); j < (n); ++j) {
        if (primes[v[i] + v[j]]) {
          cout << "2\n";
          cout << v[i] << ' ' << v[j] << endl;
          return 0;
        }
      }
    }
    cout << 1 << endl;
    cout << v[0] << endl;
    return 0;
  }
  return 0;
}

```