<|instruction|>
Solve the below problem. The solution should start with an abridged problem statement. Then key observations. Then full solution based on the observations. Then C++ and Python implementations with comments.

501. Octahedron And Dominoes
Time limit per test: 0.25 second(s)
Memory limit: 262144 kilobytes
input: standard
output: standard

 Andrew has just made a breakthrough in puzzle making: he's invented a new puzzle that is both simple to understand and unlike everything that has been invented before. His new puzzle looks like an octahedron with each of its (triangular) faces split into n2 smaller triangles by 3(n-1) lines, n-1 parallel to each side of the face. Some of smaller triangles are black and some are white. The objective is to cover all white triangles with triangular dominoes without overlapping, leaving all black triangles uncovered. A triangular domino is two small triangles with one common side. Please note that a domino can cover one small triangle on a side of the octahedron and another on the adjacent side of the octahedron (in other words, you can bend your domino in the middle). You must place each domino so that it covers exactly two white small triangles (that means you can't cover some part of a triangle with one domino and the rest with another domino, for example). How many solutions does the given puzzle have?
Input
The first line of the input file contains an integer n (1 ≤ n ≤ 4). The next 2n lines describe which small triangles are black and which are white. Imagine the octahedron standing on its vertex, with one of its other vertices pointing on us. Then the small triangles can be split into horizontal layers. The topmost layer contains 4 triangles, one per each side that ends in the top vertex. The next layer contains 12 triangles, 3 per each side that ends in the top vertex, and so on. The size of each layer increases by 8 until it reaches 8n-4 in the middle, then the next layer is again of 8n-4 small triangles (here we switched from the top 4 faces to the bottom 4 faces), and then the size of each next layer is less by 8 until we reach the bottommost layer that has just 4 triangles. The octrahedron is given layer-by-layer, and each layer is started with the leftmost small triangle on the visible part of the octahedron (the four "front" faces), goes from left to right along the visible part, and then switches to the invisible part and goes back from right to left. One layer is highlighted on the picture above. Each small triangle is specified either by '.', denoting a white triangle, or by '*', denoting a black triangle.
Output
Output one integer — the number of possible coverings of white triangles of the given octahedron with triangular dominoes.
Example(s)
sample input
sample output
2
.***
........*...
************
....
2

sample input
sample output
3
....
************
....................
********************
............
****
8

<|response|>
## 1) Abridged problem statement (concise)

An octahedron has 8 triangular faces. Each face is subdivided into \(n^2\) small triangles (\(1 \le n \le 4\)). Each small triangle is either **white** (`.`) or **black** (`*`). You must cover **all white** triangles using **triangular dominoes**, where each domino covers **exactly two adjacent** small triangles (sharing an edge). Dominoes may cross an edge between two adjacent faces (“bend”).

Input gives the coloring **layer by layer** from top to bottom (there are \(2n\) layers total, sizes grow then shrink). Output the **number of domino tilings** that cover exactly all white triangles.

---

## 2) Key observations needed to solve the problem

1. **Split into two pyramids:**  
   The octahedron is two square pyramids glued at the middle (equator):
   - upper half: 4 faces, \(n\) layers
   - lower half: 4 faces, \(n\) layers  
   Any interaction between halves happens only through the equator.

2. **Layer-by-layer DP is possible because the interface is small:**  
   In a layer, triangles alternate orientation. Between consecutive layers, only specific triangles (e.g., every other position per face row) can connect downward by sharing an edge with the next layer.  
   For layer index \(r\) (0-based within a pyramid), the number of “downward connection points” per face is \(r+1\). Across 4 faces: \(4(r+1)\).  
   At the equator \(r=n-1\), the interface size is \(4n \le 16\) bits ⇒ at most \(2^{16}=65536\) states.

3. **Solve upper and lower halves independently, then “glue” by matching boundary masks:**  
   Let:
   - `UP[M]` = ways to tile upper pyramid leaving equator carry mask `M`
   - `DOWN[M]` = ways to tile lower pyramid leaving equator carry mask `M`  
   Then total answer:
   \[
   \text{ans} = \sum_M UP[M]\cdot DOWN[M]
   \]
   (Lower half is solved bottom-up by reversing its layers to reuse the same solver.)

4. **Within one layer, brute-force enumeration is feasible:**  
   Maximum layer length in a pyramid is \(4(2n-1) \le 28\) triangles (since \(n\le 4\)).  
   For each layer and incoming boundary mask, enumerate all ways to:
   - match triangles horizontally within the layer (including wrap-around around the ring)
   - or match a triangle downward into next layer (creating a bit in next boundary mask)  
   respecting black cells and already-occupied “carry-in” cells.

---

## 3) Full solution approach

### A. Parse the octahedron into two halves
- Read `2n` strings (`layers[0..2n-1]`).
- Upper pyramid rows: `upper[0..n-1] = layers[0..n-1]`
- Lower pyramid rows (processed bottom-up): `lower[0..n-1] = reverse(layers[n..2n-1])`

Each row string length matches the number of small triangles in that layer of the displayed traversal; we treat the row as a cyclic ring for horizontal adjacency.

### B. Solve one pyramid with DP on “carry masks”
We process pyramid rows `r = 0..n-1`.

**State.**
For row `r`, define a compact mask of size `4*r` (for `r>=1`) indicating which “interface positions” (those odd positions that are reached from above) are already occupied by dominos coming from the previous row.  
We convert this compact mask to a bitmask over actual positions in the row string.

**Transition.**
Given:
- row’s black pattern (`row_mask[r]`)
- incoming occupied positions from above
we must cover all remaining white triangles using:
- horizontal dominos within the row (including wrap-around between last and first)
- downward dominos into row `r+1` where possible  
Downward choices form the next compact mask of size `4*(r+1)`.

We precompute, for each row `r>=1` and each incoming compact mask, the list of all reachable next masks by DFS over the row positions.

**Initialization (row 0 special).**
Top row has exactly 4 triangles. We can brute-force its tilings independently by trying subsets of the 4 cycle edges; this yields `first_row[blocked_mask]`.

Then `dp[1][carry_mask_of_row1]` is initialized based on how row0 is tiled and which downward dominos are used into row1.

**Result.**
After processing `n` rows, we obtain an array `res[0..2^(4n)-1]` where `res[M]` is the number of ways leaving equator carry mask `M`.

### C. Combine upper and lower results
Compute `UP = solve_pyramid(upper)`, `DOWN = solve_pyramid(lower)`.  
Answer is dot product: `sum_M UP[M]*DOWN[M]`.

---

## 4) C++ implementation (detailed comments)

```cpp
#include <bits/stdc++.h>
using namespace std;

/*
  Octahedron And Dominoes (n <= 4)

  Approach:
  - Split octahedron into upper and lower pyramids (4 faces each).
  - Solve each pyramid with layer DP:
      dp[r][mask] = number of ways after finishing rows 0..r-1,
                    with incoming carry mask for row r being 'mask'.
    Carry mask size for row r is 4*r bits (r>=1). At equator: 4*n <= 16 bits.
  - For each row, enumerate all tilings consistent with:
      - black cells are blocked
      - cells occupied by carry-in are blocked
      - remaining cells must be covered by either:
          * horizontal domino in same row (adjacent indices, plus wrap-around)
          * vertical domino going to next row (creates carry-out bit)
  - Combine upper and lower by matching equator masks:
      ans = sum_M UP[M] * DOWN[M]
*/

int n;
vector<string> layers;

static inline void read_input() {
    cin >> n;
    layers.resize(2 * n);
    for (int i = 0; i < 2 * n; i++) cin >> layers[i];
}

static vector<long long> solve_pyramid(const vector<string>& rows) {
    // Precompute number of ways to tile row0 (4-cycle) given blocked pattern.
    // blocked: 4-bit mask (1 means blocked)
    int first_row[16] = {0};
    for (int blocked = 0; blocked < 16; blocked++) {
        // Choose which of 4 cycle edges are used by dominoes:
        // edge j connects j and (j+1)%4
        for (int e = 0; e < 16; e++) {
            int covered = blocked;
            bool ok = true;
            for (int j = 0; j < 4; j++) {
                if ((e >> j) & 1) {
                    int u = j, v = (j + 1) & 3;
                    if (((covered >> u) & 1) || ((covered >> v) & 1)) {
                        ok = false;
                        break;
                    }
                    covered |= (1 << u) | (1 << v);
                }
            }
            if (ok && covered == 15) first_row[blocked]++;
        }
    }

    // Convert each row string into a bitmask of black cells.
    // row_mask[r] has bit i = 1 iff rows[r][i] == '*'
    vector<int> row_mask(n, 0);
    for (int r = 0; r < n; r++) {
        for (int i = 0; i < (int)rows[r].size(); i++) {
            if (rows[r][i] == '*') row_mask[r] |= (1 << i);
        }
    }

    /*
      Mapping details:

      Each pyramid row r consists of 4 face-segments, each of length (2r+1),
      so total row length = 4*(2r+1).

      For carry masks:
      - The "carry-in" bits for row r correspond to the triangles in row r
        that can be reached from row r-1 (these are the odd positions within
        each face segment).
      - There are r such positions per face => total 4*r bits.

      For vertical moves from row r to row r+1:
      - Only even positions (0,2,4,...) within each face segment can connect down.
      - There are (r+1) such positions per face => total 4*(r+1) bits for next mask.
    */

    int bit_mapping[5][32] = {};        // maps compact carry bit -> row position index
    int bit_next_pos[5][32];            // for a row position idx, landing position in next row
    int bit_out_index[5][32];           // for a row position idx, which bit index in next carry mask
    memset(bit_next_pos, -1, sizeof(bit_next_pos));
    memset(bit_out_index, -1, sizeof(bit_out_index));

    for (int r = 0; r < n; r++) {
        int cur_len = 2 * r + 1;        // per face segment length in this row
        int nxt_len = 2 * r + 3;        // per face segment length in next row
        int base_cur = 0;
        int base_nxt = 0;
        int out_cnt = 0;

        for (int face = 0; face < 4; face++) {
            // even positions can go down: i=0,2,4,...
            for (int i = 0; i < cur_len; i += 2) {
                int idx = base_cur + i;
                bit_next_pos[r][idx] = base_nxt + (i + 1);
                bit_out_index[r][idx] = out_cnt++;
            }
            // carry-in bits correspond to odd positions: 1,3,5,... (there are r of them)
            for (int i = 0; i < r; i++) {
                bit_mapping[r][face * r + i] = base_cur + (2 * i + 1);
            }
            base_cur += cur_len;
            base_nxt += nxt_len;
        }
    }

    // Convert compact carry mask (length 4*r) into bitmask over actual row positions.
    auto carry_to_rowmask = [&](int r, int carry_mask, int carry_bits) -> int {
        int res = 0;
        for (int b = 0; b < carry_bits; b++) {
            if ((carry_mask >> b) & 1) {
                res |= (1 << bit_mapping[r][b]);
            }
        }
        return res;
    };

    // Precompute transitions for rows r=1..n-1:
    // next_masks[r][incoming] = list of all possible outgoing carry masks for row r+1
    vector<vector<vector<int>>> next_masks(n);
    for (int r = 1; r < n; r++) {
        int in_bits = 4 * r;
        int row_len = 4 * (2 * r + 1);
        next_masks[r].assign(1 << in_bits, {});

        // DFS enumeration over positions in this row
        function<void(int,int,int,int,vector<int>&)> dfs =
            [&](int mask, int idx, int next_black, int out_mask, vector<int>& store) {
                if (idx == row_len) {
                    store.push_back(out_mask);
                    return;
                }
                if ((mask >> idx) & 1) {
                    dfs(mask, idx + 1, next_black, out_mask, store);
                    return;
                }

                // Option A: wrap-around horizontal domino (idx==0 paired with last)
                if (idx == 0 && ((mask >> (row_len - 1)) & 1) == 0) {
                    dfs(mask | (1 << (row_len - 1)), idx + 1, next_black, out_mask, store);
                }

                // Option B: vertical domino downwards (only allowed for certain idx)
                int out_i = bit_out_index[r][idx];
                if (out_i != -1) {
                    int land = bit_next_pos[r][idx];
                    if (((next_black >> land) & 1) == 0) {
                        dfs(mask, idx + 1, next_black, out_mask | (1 << out_i), store);
                    }
                }

                // Option C: horizontal domino with idx+1
                if (idx + 1 < row_len && ((mask >> (idx + 1)) & 1) == 0) {
                    dfs(mask, idx + 2, next_black, out_mask, store);
                }
            };

        for (int incoming = 0; incoming < (1 << in_bits); incoming++) {
            int occupied = carry_to_rowmask(r, incoming, in_bits);
            // If carry-in tries to occupy a black cell => impossible
            if (occupied & row_mask[r]) continue;

            int start_mask = occupied | row_mask[r];   // blocked + already occupied
            int next_black = row_mask[r + 1];          // forbid vertical landing on black
            auto &store = next_masks[r][incoming];
            dfs(start_mask, 0, next_black, 0, store);
        }
    }

    // DP array: dp[r][mask], r from 1..n, mask up to 16 bits.
    // dp[1] corresponds to carry mask into row 1 (4 bits), after fully handling row 0.
    static long long dp[5][1 << 16];
    for (int i = 0; i < 5; i++) for (int j = 0; j < (1<<16); j++) dp[i][j] = 0;

    // Initialize from row 0
    // carry into row1 has 4 bits (because r=1 => 4*r = 4)
    for (int carry = 0; carry < (1 << 4); carry++) {
        // in this implementation, carry bits for row1 also correspond to blocking pattern for row0
        // (same 4 triangles), so ensure we don't "use" a black triangle on row0 as carry.
        if (carry & row_mask[0]) continue;

        // also ensure that the implied occupied positions in row1 are not black
        if (n > 1) {
            int occupied_row1 = carry_to_rowmask(1, carry, 4);
            if (occupied_row1 & row_mask[1]) continue;
        }

        // first_row expects a 4-bit blocked mask for row0
        dp[1][carry] += first_row[carry | row_mask[0]];
    }

    // Process rows 1..n-1
    for (int r = 1; r < n; r++) {
        int in_bits = 4 * r;
        int out_bits = 4 * (r + 1);
        for (int mask = 0; mask < (1 << in_bits); mask++) {
            long long ways = dp[r][mask];
            if (!ways) continue;
            for (int nxt : next_masks[r][mask]) {
                dp[r + 1][nxt] += ways;
            }
        }
    }

    // Return result vector indexed by equator mask of size 4n bits.
    vector<long long> res(1 << (4 * n), 0);
    for (int m = 0; m < (1 << (4 * n)); m++) res[m] = dp[n][m];
    return res;
}

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

    read_input();

    vector<string> upper(n), lower(n);
    for (int i = 0; i < n; i++) upper[i] = layers[i];
    for (int i = 0; i < n; i++) lower[i] = layers[2 * n - 1 - i]; // reverse lower half

    vector<long long> up = solve_pyramid(upper);
    vector<long long> down = solve_pyramid(lower);

    long long ans = 0;
    int S = 1 << (4 * n);
    for (int m = 0; m < S; m++) ans += up[m] * down[m];

    cout << ans << "\n";
    return 0;
}
```

---

## 5) Python implementation (detailed comments)

```python
import sys
from typing import List

# Same algorithm as C++:
# - solve upper pyramid and reversed lower pyramid
# - layer DP with bitmask carries (<= 16 bits)
# - enumerate all tilings of each row via DFS recursion


def solve_pyramid(rows: List[str], n: int) -> List[int]:
    # Precompute first_row[blocked] for row0 (4-cycle)
    first_row = [0] * 16
    for blocked in range(16):
        for e in range(16):  # choose subset of the 4 cycle edges
            covered = blocked
            ok = True
            for j in range(4):
                if (e >> j) & 1:
                    u, v = j, (j + 1) & 3
                    if (covered >> u) & 1 or (covered >> v) & 1:
                        ok = False
                        break
                    covered |= (1 << u) | (1 << v)
            if ok and covered == 15:
                first_row[blocked] += 1

    # row_mask[r] bit i = 1 if rows[r][i] == '*'
    row_mask = [0] * n
    for r in range(n):
        m = 0
        for i, ch in enumerate(rows[r]):
            if ch == '*':
                m |= 1 << i
        row_mask[r] = m

    # Build mapping arrays
    bit_mapping = [[0] * 32 for _ in range(5)]     # compact carry bit -> position in row string
    bit_next_pos = [[-1] * 32 for _ in range(5)]   # row position -> landing position in next row
    bit_out_index = [[-1] * 32 for _ in range(5)]  # row position -> output carry bit index

    for r in range(n):
        cur_len = 2 * r + 1
        nxt_len = 2 * r + 3
        base_cur = 0
        base_nxt = 0
        out_cnt = 0
        for face in range(4):
            # even positions connect down
            for i in range(0, cur_len, 2):
                idx = base_cur + i
                bit_next_pos[r][idx] = base_nxt + (i + 1)
                bit_out_index[r][idx] = out_cnt
                out_cnt += 1
            # carry-in bits correspond to odd positions
            for i in range(r):
                bit_mapping[r][face * r + i] = base_cur + (2 * i + 1)
            base_cur += cur_len
            base_nxt += nxt_len

    def carry_to_rowmask(r: int, carry_mask: int, carry_bits: int) -> int:
        """Convert compact carry mask to actual row position bitmask."""
        res = 0
        for b in range(carry_bits):
            if (carry_mask >> b) & 1:
                res |= 1 << bit_mapping[r][b]
        return res

    # Precompute transitions for rows 1..n-1
    next_masks = [None] * n
    for r in range(1, n):
        in_bits = 4 * r
        row_len = 4 * (2 * r + 1)
        trans = [[] for _ in range(1 << in_bits)]

        def dfs(mask: int, idx: int, next_black: int, out_mask: int, store: List[int]):
            if idx == row_len:
                store.append(out_mask)
                return
            if (mask >> idx) & 1:
                dfs(mask, idx + 1, next_black, out_mask, store)
                return

            # wrap-around horizontal domino (0 with last)
            if idx == 0 and ((mask >> (row_len - 1)) & 1) == 0:
                dfs(mask | (1 << (row_len - 1)), idx + 1, next_black, out_mask, store)

            # vertical domino downwards
            out_i = bit_out_index[r][idx]
            if out_i != -1:
                land = bit_next_pos[r][idx]
                if ((next_black >> land) & 1) == 0:
                    dfs(mask, idx + 1, next_black, out_mask | (1 << out_i), store)

            # horizontal domino with idx+1
            if idx + 1 < row_len and ((mask >> (idx + 1)) & 1) == 0:
                dfs(mask, idx + 2, next_black, out_mask, store)

        for incoming in range(1 << in_bits):
            occupied = carry_to_rowmask(r, incoming, in_bits)
            if occupied & row_mask[r]:
                continue  # carry-in hits black cell

            start_mask = occupied | row_mask[r]
            next_black = row_mask[r + 1]
            store = trans[incoming]
            dfs(start_mask, 0, next_black, 0, store)

        next_masks[r] = trans

    # DP: dp[r][mask], but store as list-of-lists for variable mask sizes.
    dp = [None] * (n + 1)

    # Initialize dp[1] from row0
    dp[1] = [0] * (1 << 4)
    for carry in range(1 << 4):
        if carry & row_mask[0]:
            continue
        if n > 1:
            occupied_row1 = carry_to_rowmask(1, carry, 4)
            if occupied_row1 & row_mask[1]:
                continue
        dp[1][carry] += first_row[carry | row_mask[0]]

    # Transitions for rows 1..n-1
    for r in range(1, n):
        in_bits = 4 * r
        out_bits = 4 * (r + 1)
        dp[r + 1] = [0] * (1 << out_bits)
        for mask in range(1 << in_bits):
            ways = dp[r][mask]
            if not ways:
                continue
            for nxt in next_masks[r][mask]:
                dp[r + 1][nxt] += ways

    # Result is dp[n], size 2^(4n)
    return dp[n]


def main():
    data = sys.stdin.read().strip().split()
    n = int(data[0])
    layers = data[1:]  # 2n strings

    upper = layers[:n]
    lower = list(reversed(layers[n:]))

    up = solve_pyramid(upper, n)
    down = solve_pyramid(lower, n)

    ans = sum(a * b for a, b in zip(up, down))
    print(ans)


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

---

If you want, I can also add a small explicit example showing how the “carry bits” correspond to specific positions in a row (why they are the odd indices and why vertical connections are at even indices).