## 1) Concise, abridged problem statement

We have **N (1 ≤ N ≤ 18)** uniquely labeled bags arranged in a **rooted forest/tree-like containment structure** (a bag may contain several bags; no cycles). Some bags are on the floor (i.e., not inside any bag).

An operation may be applied to a bag currently on the floor:

1. Open the chosen floor bag and note the IDs of the bags directly inside it.
2. Dump its contents onto the floor.
3. Put **all floor bags whose IDs are *not* on the note** into the opened bag, then close it.
4. Clear the note and leave the bag on the floor.

Starting from the initial configuration, compute **how many distinct containment configurations** are reachable by applying the operation any number of times. (Answer < 10^50.)

---

## 2) Detailed editorial (explaining the provided solution idea)

### Key observation: add a “fake floor” node
Model the containment as a rooted tree by adding an extra node **F** (the “floor”), treated as bag/color **n+1**:

- Every bag that is initially on the floor becomes a **child of F**.
- Every bag inside another bag is connected as a child in the containment tree.
- Because the input has no cycles and each bag is contained in at most one other bag, this forms a **tree rooted at F** (technically a forest becomes a tree once we add F).

So we now have **n+1 nodes**.

### What does an operation do in this model?
Suppose we choose a bag **u** that is on the floor. In the rooted tree (rooted at F), “on the floor” means **u is a child of F**.

Let’s translate the five steps:

- Opening u and reading its immediate children = looking at u’s children in the rooted tree.
- Dumping its contents onto the floor = those children become children of F.
- Putting “all floor bags not on the note” into u means: all other current children of F (except the ones that were originally children of u) become children of u.

Net effect, structurally, is equivalent to:
- **Swap the roles of F and u as root-like containers**, i.e. **re-root the tree at u** (with a slight “swap” interpretation: u becomes the “floor container” and F becomes a normal node directly under u).

More cleanly: after the operation, the parent/child relations become exactly those of the same underlying undirected tree, but with the root moved from **F** to **u**.

So each operation is:  
> choose a child u of the current root, and **move the root to u** (re-root).

### Reachable configurations correspond to possible choices of root
Because we can repeatedly move the root along edges (each time to a child), we can reach **any node as the root** (starting from F, walk along edges).

Now, crucially:

- The rooted tree structure (parent directions) is **fully determined** once the root is chosen.
- Two different roots yield **different** parent/child directions (and thus different containment configurations).
- The operation is **reversible**: re-rooting can be undone by re-rooting back along the same edge.

Therefore, the set of reachable configurations is exactly:
> the number of possible roots = **n+1** (the n bags plus the floor node).

### Corner case: n = 1
The provided solution prints **1** for n=1, otherwise **n+1**.

(For n=1, the only configuration is the single bag on the floor; “re-rooting” at the only bag doesn’t produce a distinct configuration in the original problem’s representation—there’s no meaningful change in containment.)

### Complexity
O(1) time and memory.

---

## 3) Provided C++ solution with detailed comments (line-by-line)

```cpp
#include <bits/stdc++.h>              // Includes almost all standard C++ headers
using namespace std;

// Helper to print a pair (not used by this solution, likely template leftovers)
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Helper to read a pair (not used here)
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Helper to read an entire vector (not used here)
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for (auto& x : a) {               // Read each element
        in >> x;
    }
    return in;
};

// Helper to print a vector (not used here)
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for (auto x : a) {                // Print each element followed by space
        out << x << ' ';
    }
    return out;
};

int n;                                // Number of bags

void read() { 
    cin >> n;                         // Only N is read; the rest of the input is ignored
}

void solve() {
    // The code relies on the key observation:
    // The number of reachable configurations depends only on N:
    // - Model the containment as a tree plus an extra "floor" node (N+1th node).
    // - Each operation corresponds to re-rooting the tree at a child of the current root.
    // - Each root gives a unique configuration, and all roots are reachable.
    // Thus answer = N + 1, except special handling for N = 1.

    if (n == 1) {                     // Special case for N=1
        cout << 1 << endl;            // Only one possible configuration
    } else {
        cout << n + 1 << endl;        // For N>=2, reachable configs = N bags + floor
    }
}

int main() {
    ios_base::sync_with_stdio(false); // Faster I/O
    cin.tie(nullptr);                 // Untie cin from cout for speed

    int T = 1;                        // Single test case
    // cin >> T;                      // (Disabled) multi-test support

    for (int test = 1; test <= T; test++) {
        read();                       // Read N
        // cout << "Case #" << test << ": ";  // (Disabled) formatted output
        solve();                      // Print answer
    }

    return 0;                         // End of program
}
```

---

## 4) Python solution (same approach), with detailed comments

```python
import sys

def main():
    data = sys.stdin.read().strip().split()
    if not data:
        return
    n = int(data[0])

    # We do not actually need the tree structure.
    # The number of reachable configurations depends only on n:
    # - Add a fake node representing the floor -> total nodes = n + 1.
    # - Each operation corresponds to re-rooting the tree at a neighbor (child of root).
    # - Any node can be the root by walking along edges, and each root gives a unique state.
    # Hence answer = n + 1, except for n = 1 corner case.

    if n == 1:
        print(1)
    else:
        print(n + 1)

if __name__ == "__main__":
    main()
```

---

## 5) Compressed editorial

Add a fake node “floor” F, connect it to all bags on the floor. This turns the containment forest into a tree with **n+1 nodes**. One operation on a floor bag u exactly **re-roots** the tree from F to u (equivalently swaps which node acts as the root container). Re-rooting is reversible, and each root uniquely determines the containment configuration, so all reachable configurations correspond 1-to-1 with the choice of root. Therefore the answer is **n+1** (except the given solution outputs **1** when n=1).