#include <bits/stdc++.h>
using namespace std;
bitset<2000005> bs;
vector<int> p;
void build() {
  bs[0] = bs[1] = 1;
  for (int i = 2; i < 2000005; i++) {
    if (!bs[i]) p.push_back(i);
    for (int j = 0; i * p[j] < 2000005; j++) {
      bs[i * p[j]] = 1;
      if (i % p[j] == 0) break;
    }
  }
}
int in[1005];
bitset<1005> G[1005], now;
int main() {
  {
    cin.tie(0);
    ios_base::sync_with_stdio(false);
  };
  int tt = clock();
  build();
  int n;
  cin >> n;
  vector<int> v;
  for (int i = 0; i < n; i++) cin >> in[i], v.push_back(i);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i != j && !bs[in[i] + in[j]]) {
        G[i][j] = G[j][i] = 1;
      }
  int ans = 0;
  bitset<1005> best;
  while ((clock() - tt) * 1.0 / CLOCKS_PER_SEC < 0.95) {
    random_shuffle(v.begin(), v.end());
    now.reset();
    int r = 0;
    for (int i : v)
      if ((G[i] & now) == now) {
        now[i] = 1;
        r++;
      }
    if (ans < r) {
      ans = r;
      best = now;
    }
  }
  cout << ans << '\n';
  for (int i = 0; i < n; i++)
    if (best[i]) cout << in[i] << ' ';
  return 0;
}

### Prompt

Generate 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;
bitset<2000005> bs;
vector<int> p;
void build() {
  bs[0] = bs[1] = 1;
  for (int i = 2; i < 2000005; i++) {
    if (!bs[i]) p.push_back(i);
    for (int j = 0; i * p[j] < 2000005; j++) {
      bs[i * p[j]] = 1;
      if (i % p[j] == 0) break;
    }
  }
}
int in[1005];
bitset<1005> G[1005], now;
int main() {
  {
    cin.tie(0);
    ios_base::sync_with_stdio(false);
  };
  int tt = clock();
  build();
  int n;
  cin >> n;
  vector<int> v;
  for (int i = 0; i < n; i++) cin >> in[i], v.push_back(i);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i != j && !bs[in[i] + in[j]]) {
        G[i][j] = G[j][i] = 1;
      }
  int ans = 0;
  bitset<1005> best;
  while ((clock() - tt) * 1.0 / CLOCKS_PER_SEC < 0.95) {
    random_shuffle(v.begin(), v.end());
    now.reset();
    int r = 0;
    for (int i : v)
      if ((G[i] & now) == now) {
        now[i] = 1;
        r++;
      }
    if (ans < r) {
      ans = r;
      best = now;
    }
  }
  cout << ans << '\n';
  for (int i = 0; i < n; i++)
    if (best[i]) cout << in[i] << ' ';
  return 0;
}

```