#include <bits/stdc++.h>
using namespace std;
long long lp[20000002];
void cal() {
  for (long long i = 2; i <= 20000000; i++) {
    if (lp[i] == 0) {
      lp[i] = i;
      for (long long j = i; j <= 20000000; j += i) {
        lp[j] = i;
      }
    }
  }
}
int main() {
  cal();
  long long i, j, k, l, m, n;
  cin >> n;
  vector<long long> a, b;
  long long bt = 0;
  for (i = 0; i < n; i++) {
    cin >> l;
    if (l == 1) {
      bt++;
    }
    if (l % 2) {
      a.push_back(l);
    } else
      b.push_back(l);
  }
  if (bt < 2) bt = 0;
  vector<long long> v;
  for (auto x : a) {
    for (auto xx : b) {
      if (lp[x + xx] == (x + xx)) {
        if (x == 1) {
          if (bt) {
            v.clear();
            for (i = 0; i < bt; i++) v.push_back(x);
            v.push_back(xx);
          } else if (v.empty()) {
            v.push_back(x);
            v.push_back(xx);
          }
        } else if (v.empty()) {
          v.push_back(x);
          v.push_back(xx);
        }
      }
    }
  }
  for (auto x : a) {
    if (lp[x] == x) {
      if (v.empty()) v.push_back(x);
    }
    if (x == 1) {
      if (v.size() < bt) {
        v.clear();
        for (i = 0; i < bt; i++) v.push_back(x);
      }
    }
  }
  for (auto xx : b) {
    if (xx == 2) {
      if (v.empty()) v.push_back(xx);
    }
    if (v.size()) break;
  }
  if (v.empty()) {
    if (a.size())
      v.push_back(a[0]);
    else
      v.push_back(b[0]);
  }
  cout << v.size() << endl;
  for (auto x : v) cout << x << " ";
  cout << endl;
}

### Prompt

Please formulate 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;
long long lp[20000002];
void cal() {
  for (long long i = 2; i <= 20000000; i++) {
    if (lp[i] == 0) {
      lp[i] = i;
      for (long long j = i; j <= 20000000; j += i) {
        lp[j] = i;
      }
    }
  }
}
int main() {
  cal();
  long long i, j, k, l, m, n;
  cin >> n;
  vector<long long> a, b;
  long long bt = 0;
  for (i = 0; i < n; i++) {
    cin >> l;
    if (l == 1) {
      bt++;
    }
    if (l % 2) {
      a.push_back(l);
    } else
      b.push_back(l);
  }
  if (bt < 2) bt = 0;
  vector<long long> v;
  for (auto x : a) {
    for (auto xx : b) {
      if (lp[x + xx] == (x + xx)) {
        if (x == 1) {
          if (bt) {
            v.clear();
            for (i = 0; i < bt; i++) v.push_back(x);
            v.push_back(xx);
          } else if (v.empty()) {
            v.push_back(x);
            v.push_back(xx);
          }
        } else if (v.empty()) {
          v.push_back(x);
          v.push_back(xx);
        }
      }
    }
  }
  for (auto x : a) {
    if (lp[x] == x) {
      if (v.empty()) v.push_back(x);
    }
    if (x == 1) {
      if (v.size() < bt) {
        v.clear();
        for (i = 0; i < bt; i++) v.push_back(x);
      }
    }
  }
  for (auto xx : b) {
    if (xx == 2) {
      if (v.empty()) v.push_back(xx);
    }
    if (v.size()) break;
  }
  if (v.empty()) {
    if (a.size())
      v.push_back(a[0]);
    else
      v.push_back(b[0]);
  }
  cout << v.size() << endl;
  for (auto x : v) cout << x << " ";
  cout << endl;
}

```