#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e6 + 7;
const long long mod = 1e9 + 7;
const long long INF = 1e18 + 7;
const int inf = 1e9;
const long long maxx = 1e6 + 7;
int n;
int a[maxn];
bool vis[maxn];
int primer[maxn / 10];
int cnt = 0;
void find_primer() {
  for (int i = 2; i <= maxn; i++) {
    if (!vis[i]) primer[cnt++] = i;
    for (int j = 0; j < cnt && primer[j] * i <= maxn; j++) {
      vis[i * primer[j]] = 1;
      if (i % primer[j] == 0) break;
    }
  }
}
int main() {
  cin >> n;
  find_primer();
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a + 1, a + 1 + n);
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1) {
      cnt++;
    } else {
      break;
    }
  }
  int flag = 0, tmp = 1;
  for (int i = cnt + 1; i <= n; i++) {
    if (!vis[a[i] + 1]) {
      tmp = a[i];
      flag = 1;
      break;
    }
  }
  if (cnt > 0) {
    if (flag) cnt++;
  }
  int flag1 = 0;
  int ii, jj;
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (!vis[a[i] + a[j]]) {
        flag1 = 1;
        ii = i, jj = j;
        break;
      }
    }
    if (flag1) break;
  }
  if (cnt > 2) {
    cout << cnt << endl;
    for (int i = 1; i < cnt; i++) {
      cout << a[i] << ' ';
    }
    cout << tmp << endl;
  } else if (flag1) {
    cout << "2" << endl;
    cout << a[ii] << " " << a[jj] << endl;
  } else {
    puts("1");
    cout << a[1] << endl;
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 maxn = 2e6 + 7;
const long long mod = 1e9 + 7;
const long long INF = 1e18 + 7;
const int inf = 1e9;
const long long maxx = 1e6 + 7;
int n;
int a[maxn];
bool vis[maxn];
int primer[maxn / 10];
int cnt = 0;
void find_primer() {
  for (int i = 2; i <= maxn; i++) {
    if (!vis[i]) primer[cnt++] = i;
    for (int j = 0; j < cnt && primer[j] * i <= maxn; j++) {
      vis[i * primer[j]] = 1;
      if (i % primer[j] == 0) break;
    }
  }
}
int main() {
  cin >> n;
  find_primer();
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  sort(a + 1, a + 1 + n);
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1) {
      cnt++;
    } else {
      break;
    }
  }
  int flag = 0, tmp = 1;
  for (int i = cnt + 1; i <= n; i++) {
    if (!vis[a[i] + 1]) {
      tmp = a[i];
      flag = 1;
      break;
    }
  }
  if (cnt > 0) {
    if (flag) cnt++;
  }
  int flag1 = 0;
  int ii, jj;
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (!vis[a[i] + a[j]]) {
        flag1 = 1;
        ii = i, jj = j;
        break;
      }
    }
    if (flag1) break;
  }
  if (cnt > 2) {
    cout << cnt << endl;
    for (int i = 1; i < cnt; i++) {
      cout << a[i] << ' ';
    }
    cout << tmp << endl;
  } else if (flag1) {
    cout << "2" << endl;
    cout << a[ii] << " " << a[jj] << endl;
  } else {
    puts("1");
    cout << a[1] << endl;
  }
}

```