#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 228;
int n, a[N], mrk[N], kol[N], ans, a1, a2, t;
void up1(int x) {
  if (kol[x] > ans) {
    ans = kol[x];
    t = 1;
    a1 = x;
  }
}
void up2(int x, int y) {
  if (kol[x] + kol[y] > ans) {
    ans = kol[x] + kol[y];
    a1 = x;
    a2 = y;
    t = 2;
  }
}
void up3(int x, int y) {
  swap(x, y);
  if (kol[x] + 1 > ans) {
    ans = kol[x] + 1;
    a1 = x;
    a2 = y;
    t = 3;
  }
}
void up4(int x, int y) {
  if (2 > ans) {
    ans = 2;
    a1 = x;
    a2 = y;
    t = 4;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  for (int i = 2; i <= 2e6; i++) {
    if (mrk[i]) continue;
    for (int j = 2; i * j <= 2e6; j++) mrk[i * j] = 1;
  }
  cin >> n;
  ans = 0;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) kol[a[i]]++;
  for (int i = 1; i <= n; i++)
    if (!mrk[a[i]] && !mrk[a[i] + a[i]]) up1(a[i]);
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++) {
      if (a[i] == a[j]) continue;
      if (!mrk[a[i] + a[j]] && !mrk[a[i] + a[i]] && !mrk[a[j] + a[j]])
        up2(a[i], a[j]);
      if (!mrk[a[i] + a[j]] && mrk[a[i] + a[i]] && !mrk[a[j] + a[j]])
        up3(a[i], a[j]);
      if (!mrk[a[i] + a[j]] && !mrk[a[i] + a[i]] && mrk[a[j] + a[j]])
        up3(a[j], a[i]);
      if (!mrk[a[i] + a[j]] && mrk[a[i] + a[i]] && mrk[a[j] + a[j]])
        up4(a[j], a[i]);
    }
  if (t == 0) {
    cout << 1 << endl;
    cout << a[1] << endl;
    return 0;
  }
  cout << ans << endl;
  if (t == 3) {
    for (int i = 1; i <= n; i++)
      if (a[i] == a1) cout << a[i] << " ";
    cout << a2 << endl;
    return 0;
  }
  if (t == 4) {
    cout << a1 << " " << a2 << endl;
    return 0;
  }
  if (t == 1) {
    for (int i = 1; i <= n; i++)
      if (a[i] == a1) cout << a[i] << " ";
    return 0;
  }
  for (int i = 1; i <= n; i++)
    if (a[i] == a1 || a[i] == a2) cout << a[i] << " ";
  return 0;
}

### 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;
const int N = 2e6 + 228;
int n, a[N], mrk[N], kol[N], ans, a1, a2, t;
void up1(int x) {
  if (kol[x] > ans) {
    ans = kol[x];
    t = 1;
    a1 = x;
  }
}
void up2(int x, int y) {
  if (kol[x] + kol[y] > ans) {
    ans = kol[x] + kol[y];
    a1 = x;
    a2 = y;
    t = 2;
  }
}
void up3(int x, int y) {
  swap(x, y);
  if (kol[x] + 1 > ans) {
    ans = kol[x] + 1;
    a1 = x;
    a2 = y;
    t = 3;
  }
}
void up4(int x, int y) {
  if (2 > ans) {
    ans = 2;
    a1 = x;
    a2 = y;
    t = 4;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  for (int i = 2; i <= 2e6; i++) {
    if (mrk[i]) continue;
    for (int j = 2; i * j <= 2e6; j++) mrk[i * j] = 1;
  }
  cin >> n;
  ans = 0;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) kol[a[i]]++;
  for (int i = 1; i <= n; i++)
    if (!mrk[a[i]] && !mrk[a[i] + a[i]]) up1(a[i]);
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++) {
      if (a[i] == a[j]) continue;
      if (!mrk[a[i] + a[j]] && !mrk[a[i] + a[i]] && !mrk[a[j] + a[j]])
        up2(a[i], a[j]);
      if (!mrk[a[i] + a[j]] && mrk[a[i] + a[i]] && !mrk[a[j] + a[j]])
        up3(a[i], a[j]);
      if (!mrk[a[i] + a[j]] && !mrk[a[i] + a[i]] && mrk[a[j] + a[j]])
        up3(a[j], a[i]);
      if (!mrk[a[i] + a[j]] && mrk[a[i] + a[i]] && mrk[a[j] + a[j]])
        up4(a[j], a[i]);
    }
  if (t == 0) {
    cout << 1 << endl;
    cout << a[1] << endl;
    return 0;
  }
  cout << ans << endl;
  if (t == 3) {
    for (int i = 1; i <= n; i++)
      if (a[i] == a1) cout << a[i] << " ";
    cout << a2 << endl;
    return 0;
  }
  if (t == 4) {
    cout << a1 << " " << a2 << endl;
    return 0;
  }
  if (t == 1) {
    for (int i = 1; i <= n; i++)
      if (a[i] == a1) cout << a[i] << " ";
    return 0;
  }
  for (int i = 1; i <= n; i++)
    if (a[i] == a1 || a[i] == a2) cout << a[i] << " ";
  return 0;
}

```