#include <bits/stdc++.h>
using namespace std;
template <class T>
inline T bigmod(T p, T e, T M) {
  if (e == 0) return 1;
  if (e % 2 == 0) {
    T t = bigmod(p, e / 2, M);
    return (t * t) % M;
  }
  return (bigmod(p, e - 1, M) * p) % M;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
template <class T>
inline T multimod(T a, T b, T m) {
  T ans = 0ll;
  a %= m, b %= m;
  while (b) {
    if (b & 1ll) ans = m - ans > a ? (ans + a) : (ans + a - m);
    b >>= 1ll;
    a = (m - a) > a ? a + a : a + a - m;
  }
  return (T)ans;
}
bool is[(2000000 + 7)];
void sieve() {
  int sq = sqrt((2000000 + 7));
  is[1] = 1;
  int i, j;
  for (i = 4; i < (2000000 + 7); i += 2) is[i] = 1;
  for (i = 3; i <= sq; i += 2) {
    if (!is[i]) {
      for (j = i + i; j < (2000000 + 7); j += i) is[j] = 1;
    }
  }
}
int arr[(2000000 + 7)];
int main() {
  sieve();
  int i, j, k;
  int n, num, m, q;
  cin >> n;
  int cnt = 0;
  for (i = 0; i < n; i++) {
    cin >> arr[i];
    if (arr[i] == 1) cnt++;
  }
  if (cnt >= 2) {
    bool ck = 0;
    for (i = 0; i < n; i++) {
      if (arr[i] == 1) continue;
      if (is[arr[i] + 1] == 0) {
        cout << cnt + 1 << endl;
        cout << arr[i] << " ";
        ck = 1;
        break;
      }
    }
    if (!ck) cout << cnt << endl;
    for (i = 0; i < cnt; i++) cout << 1 << " ";
    return 0;
  } else {
    for (i = 0; i < n; i++) {
      for (j = i + 1; j < n; j++) {
        if (!is[arr[i] + arr[j]]) {
          cout << 2 << endl;
          cout << arr[i] << " " << arr[j] << endl;
          return 0;
        }
      }
    }
  }
  cout << 1 << endl;
  cout << arr[0] << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
template <class T>
inline T bigmod(T p, T e, T M) {
  if (e == 0) return 1;
  if (e % 2 == 0) {
    T t = bigmod(p, e / 2, M);
    return (t * t) % M;
  }
  return (bigmod(p, e - 1, M) * p) % M;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
template <class T>
inline T multimod(T a, T b, T m) {
  T ans = 0ll;
  a %= m, b %= m;
  while (b) {
    if (b & 1ll) ans = m - ans > a ? (ans + a) : (ans + a - m);
    b >>= 1ll;
    a = (m - a) > a ? a + a : a + a - m;
  }
  return (T)ans;
}
bool is[(2000000 + 7)];
void sieve() {
  int sq = sqrt((2000000 + 7));
  is[1] = 1;
  int i, j;
  for (i = 4; i < (2000000 + 7); i += 2) is[i] = 1;
  for (i = 3; i <= sq; i += 2) {
    if (!is[i]) {
      for (j = i + i; j < (2000000 + 7); j += i) is[j] = 1;
    }
  }
}
int arr[(2000000 + 7)];
int main() {
  sieve();
  int i, j, k;
  int n, num, m, q;
  cin >> n;
  int cnt = 0;
  for (i = 0; i < n; i++) {
    cin >> arr[i];
    if (arr[i] == 1) cnt++;
  }
  if (cnt >= 2) {
    bool ck = 0;
    for (i = 0; i < n; i++) {
      if (arr[i] == 1) continue;
      if (is[arr[i] + 1] == 0) {
        cout << cnt + 1 << endl;
        cout << arr[i] << " ";
        ck = 1;
        break;
      }
    }
    if (!ck) cout << cnt << endl;
    for (i = 0; i < cnt; i++) cout << 1 << " ";
    return 0;
  } else {
    for (i = 0; i < n; i++) {
      for (j = i + 1; j < n; j++) {
        if (!is[arr[i] + arr[j]]) {
          cout << 2 << endl;
          cout << arr[i] << " " << arr[j] << endl;
          return 0;
        }
      }
    }
  }
  cout << 1 << endl;
  cout << arr[0] << endl;
  return 0;
}

```