## 1) Abridged problem statement

There is a 14×4 patience layout, but we consider a restricted family of positions:

- Three suits are already completely ordered (Ace…King), hence irrelevant.
- The remaining suit has its left part already ordered, and only the last `n` columns (rightmost positions) may be "unordered" among the remaining high cards plus one empty cell (column 14 must be empty in the goal).
- A move: pick an empty cell. Look at its left neighbor. If it contains a card of value `v` (not King), you must move the next higher card of the same suit (`v+1`) into the empty cell, creating a new empty where that card came from. If the left neighbor is King or empty, no move.

For a given `n (1..13)`, count how many such restricted positions have a winning strategy (can reach the fully ordered row with the empty at the end).

---

## 2) Detailed editorial (what the solution is doing)

### Key reduction: only one suit matters, and only `n` "tokens"
Because 3 suits are fully solved, the game reduces to a single row (the remaining suit) with:

- A fixed ordered prefix (left part).
- A suffix of length `n` that contains the remaining highest cards of that suit plus one empty cell in some order.

Example for `n=3`: the suffix consists of `{Q, K, empty}` arranged in any of 3! = 6 ways.

So the state is just a permutation of `n` distinct symbols:
- Cards: consecutive ranks near the top (e.g. `13-n+1 .. 13` if King=13)
- One empty slot

### Move rule in the reduced model
Label the ordered prefix ends with some rank `r` (the last correctly placed card). Then:

- If a position (slot) in the suffix is empty, and its left neighbor contains rank `x` (a card), and `x` is not King, then the card `x+1` (which is somewhere in the suffix) can be moved into the empty slot.

That means a move is fully determined by the location of the empty slot and the rank immediately to its left.

### Winning positions = ancestors of the goal in the predecessor graph
Let `G` be the goal arrangement (fully ordered, empty at the end).

A state is winning iff there exists a sequence of moves leading to `G`. In graph terms: the winning set is exactly the set of nodes that can reach `G`.

Instead of exploring forward from every state, do the classic trick:

- Build the set of all states that can reach the goal by exploring backwards:
  - Start with `{G}`.
  - Repeatedly add any state that has a legal move into a state already known to be winning.
  - Use a visited set to avoid processing states twice.

Finally, the size of this set is the answer for that `n`.

### Complexity and why the submitted code uses precomputation
A direct brute force over states runs in roughly O(n!), which for `n=13` is on the order of billions of operations and a reachable set growing to ~118 million — far too slow for the 0.25s limit if done online.

But there are only 13 possible inputs (`n=1..13`), so you can:
- Run the brute force offline once,
- Record answers for all `n`,
- Submit a program that just prints the precomputed value.

That's exactly what the provided C++ does.

### The precomputed answers
The code stores:

```
n : answer
1 : 1
2 : 2
3 : 5
4 : 14
5 : 47
6 : 189
7 : 891
8 : 4815
9 : 29547
10: 203173
11: 1548222
12: 12966093
13: 118515434
```

(And a dummy 0 at index 0 for convenience.)

---

## 3) C++ Solution

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n;

void read() { cin >> n; }

void solve() {
    // The solution here is to simply simulate this. We can create a brute force
    // that runs in O(n!), which for n=13 is 6.2B operations, or on modern
    // computers will take less than a minute. Then we precompute the answer for
    // all possible values of n - after all there aren't that many. In terms of
    // the brute force, the smart way is to build the set of all ancestors of
    // the goal in the predecessor graph. Then starting from the goal, we
    // repeatedly add every valid predecessor (states that can move forward into
    // the current set). It's useful to keep a set of states we have visited to
    // avoid duplicates, as this cuts the runtime. The size of the reached set
    // is exactly the answer. Overall, the size of the set grows up to ~118M, so
    // something like this could in theory pass if optimized.

    static const array<int64_t, 14> answers = {
        0ll,   1ll,    2ll,     5ll,      14ll,      47ll,       189ll,
        891ll, 4815ll, 29547ll, 203173ll, 1548222ll, 12966093ll, 118515434ll
    };

    cout << answers[n] << endl;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}
```

---

## 4) Python solution (same approach: output precomputed answers)

```python
import sys

def main() -> None:
    # Read n (1..13). Input contains a single integer.
    data = sys.stdin.read().strip().split()
    n = int(data[0])

    # Precomputed answers for each n.
    # Index 0 is a dummy so we can access answers[n] directly.
    answers = [
        0,
        1,
        2,
        5,
        14,
        47,
        189,
        891,
        4815,
        29547,
        203173,
        1548222,
        12966093,
        118515434
    ]

    # Output the precomputed number of winning positions.
    sys.stdout.write(str(answers[n]))

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

---

## 5) Compressed editorial

Reduce to the last `n` columns of the only unsolved suit: a finite permutation state with one empty cell and several highest cards. A move is possible only at the empty cell: if its left neighbor is rank `x` (not King), move rank `x+1` into the empty. Winning states are exactly those that can reach the goal ordering, i.e., the backward-reachable set from the goal in the predecessor graph. For `n` up to 13 this can be brute-forced offline; answers are then hardcoded due to very tight time limit. The submitted solution simply prints the precomputed value for the given `n`.
