#include <bits/stdc++.h>
using namespace std;
const int INF = (int)(1e9);
const long long INF64 = (long long)(1e18);
const long double EPS = 1e-9;
const long double PI = 3.1415926535897932384626433832795;
const int N = 2000002;
int n;
bool prime[2000005];
void sieve() {
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= N; p++) {
    if (prime[p] == true) {
      for (int i = p * 2; i <= N; i += p) prime[i] = false;
    }
  }
  if (prime[4] == true) printf("Yes\n");
}
int main() {
  sieve();
  scanf("%d", &n);
  int arr[1005], countOnes = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &arr[i]);
    if (arr[i] == 1) countOnes++;
  }
  int primeWithOnes;
  int primeWithOnesCount = 0;
  vector<int> twoSumPrimes;
  if (countOnes) {
    for (int i = 0; i < n; ++i) {
      if (arr[i] != 1) {
        if (prime[arr[i] + 1] == true) {
          primeWithOnes = arr[i];
          primeWithOnesCount++;
          break;
        }
      }
    }
  }
  int flag = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      if (prime[arr[i] + arr[j]] == true) {
        twoSumPrimes.push_back(arr[i]);
        twoSumPrimes.push_back(arr[j]);
        flag = 1;
        break;
      }
    }
    if (flag == 1) break;
  }
  int singlePrime;
  for (int i = 0; i < n; ++i) {
    if (prime[arr[i]] == true) {
      singlePrime = arr[i];
      break;
    }
  }
  if (countOnes + primeWithOnesCount > twoSumPrimes.size() &&
      countOnes + primeWithOnesCount > 1) {
    printf("%d\n", countOnes + primeWithOnesCount);
    for (int i = 0; i < countOnes; ++i) printf("1 ");
    if (primeWithOnesCount != 0) printf("%d ", primeWithOnes);
    printf("\n");
  } else if (twoSumPrimes.size() > 1) {
    printf("%d\n", (int)twoSumPrimes.size());
    for (int i = 0; i < twoSumPrimes.size(); ++i)
      printf("%d ", (int)twoSumPrimes[i]);
    printf("\n");
  } else {
    printf("1\n");
    printf("%d\n", singlePrime);
  }
  return 0;
}

### Prompt

Your task is to create 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 INF = (int)(1e9);
const long long INF64 = (long long)(1e18);
const long double EPS = 1e-9;
const long double PI = 3.1415926535897932384626433832795;
const int N = 2000002;
int n;
bool prime[2000005];
void sieve() {
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= N; p++) {
    if (prime[p] == true) {
      for (int i = p * 2; i <= N; i += p) prime[i] = false;
    }
  }
  if (prime[4] == true) printf("Yes\n");
}
int main() {
  sieve();
  scanf("%d", &n);
  int arr[1005], countOnes = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", &arr[i]);
    if (arr[i] == 1) countOnes++;
  }
  int primeWithOnes;
  int primeWithOnesCount = 0;
  vector<int> twoSumPrimes;
  if (countOnes) {
    for (int i = 0; i < n; ++i) {
      if (arr[i] != 1) {
        if (prime[arr[i] + 1] == true) {
          primeWithOnes = arr[i];
          primeWithOnesCount++;
          break;
        }
      }
    }
  }
  int flag = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j) {
      if (prime[arr[i] + arr[j]] == true) {
        twoSumPrimes.push_back(arr[i]);
        twoSumPrimes.push_back(arr[j]);
        flag = 1;
        break;
      }
    }
    if (flag == 1) break;
  }
  int singlePrime;
  for (int i = 0; i < n; ++i) {
    if (prime[arr[i]] == true) {
      singlePrime = arr[i];
      break;
    }
  }
  if (countOnes + primeWithOnesCount > twoSumPrimes.size() &&
      countOnes + primeWithOnesCount > 1) {
    printf("%d\n", countOnes + primeWithOnesCount);
    for (int i = 0; i < countOnes; ++i) printf("1 ");
    if (primeWithOnesCount != 0) printf("%d ", primeWithOnes);
    printf("\n");
  } else if (twoSumPrimes.size() > 1) {
    printf("%d\n", (int)twoSumPrimes.size());
    for (int i = 0; i < twoSumPrimes.size(); ++i)
      printf("%d ", (int)twoSumPrimes[i]);
    printf("\n");
  } else {
    printf("1\n");
    printf("%d\n", singlePrime);
  }
  return 0;
}

```