## 1. Abridged Problem Statement

An equilateral triangle of side length `N` is divided into `N²` small unit triangles. Each unit triangle is one of 4 kinds, where kind `i` has a fixed pattern of black/white edges, and triangles may be rotated freely.

You are given counts `n1, n2, n3, n4` of the four kinds, summing to `N²`. Count how many different ways to assemble the big triangle so that every pair of adjacent unit triangles has matching colors on their shared edge.

The big triangle’s orientation is fixed, so rotated whole-triangle colorings are considered different.

Constraints:

```text
1 ≤ N ≤ 5
n1 + n2 + n3 + n4 = N²
```

---

## 2. Detailed Editorial

### Geometry of the board

A triangle of side `N` contains `N²` unit triangles.

We process the board row by row. Row `r`, using 1-based indexing, contains:

```text
2r - 1
```

unit triangles.

Positions inside a row are indexed by `p`, from `0` to `2r - 2`.

The orientation alternates:

- even `p`: upward-facing triangle
- odd `p`: downward-facing triangle

So row `r` looks like:

```text
up, down, up, down, ..., up
```

---

### Edge representation

Each small triangle has 3 edges. The solution represents their colors as bits:

```text
0 = white
1 = black
```

For a triangle, the three relevant sides are stored as:

```text
left, right, third
```

where:

- for an upward triangle, `third` is its bottom side
- for a downward triangle, `third` is its top side

The four triangle kinds correspond to having:

```text
kind 0: 0 black edges
kind 1: 1 black edge
kind 2: 2 black edges
kind 3: 3 black edges
```

The input gives counts of kinds `1..4`, but internally the code stores them as `cnt[0..3]`.

For a triangle of kind `k`, we enumerate every 3-bit mask with exactly `k` bits set. This corresponds to all possible rotations/orientations of that triangle kind.

---

### Shared edges

There are two types of edges shared between neighboring cells.

#### 1. Vertical-ish edges inside the same row

`v_edge[r][p]` stores the edge between cells `p` and `p + 1` of row `r`.

It is:

- the right side of cell `(r, p)`
- the left side of cell `(r, p + 1)`

#### 2. Horizontal edges between rows

`h_edge[r][k]` stores the edge between row `r` and row `r + 1`.

It is:

- the bottom side of upward cell `(r, 2k)`
- the top side of downward cell `(r + 1, 2k + 1)`

---

### DFS with memoization

We process cells in row-major order.

At each cell `(r, p)`, some edges may already be fixed by previously placed triangles:

- If `p > 0`, the left edge is fixed by the previous cell in the same row.
- If the current triangle is downward-facing, its top edge is fixed by a triangle in the previous row.

For each available triangle kind, we try every orientation, keeping only those that match already-fixed neighboring edges.

Then we store the newly determined outgoing edges:

- the right edge into `v_edge[r][p]`
- if upward-facing, the bottom edge into `h_edge[r][p / 2]`

Then recurse to the next cell.

---

### Why memoization is possible

Naively, the search tree is huge. However, at any point, most already-written edges will never be used again.

The future only depends on:

1. Current cell position `(r, p)`
2. Remaining counts of the four triangle kinds
3. A small active boundary of edge colors that future cells still need

This active boundary consists of:

- unconsumed horizontal edges from the previous row
- the immediate left edge of the current cell
- already built horizontal edges from the current row that the next row will need

The function `active_mask(r, p)` packs these still-relevant edge colors into a bitmask.

Then memoization key is:

```text
(r, p, active_mask, cnt[0], cnt[1], cnt[2], cnt[3])
```

Since `N ≤ 5`, the number of active edges is tiny, so this DP is very fast.

---

### Base case

When `r > N`, all cells have been filled successfully, so we return `1`.

---

### Complexity

There are only `N² ≤ 25` cells.

For each state, we try:

- 4 triangle kinds
- up to 8 edge masks

So each transition is small.

The memoized state space is also small because the active boundary has only `O(N)` bits and the counts sum to at most 25.

---

## 3. Provided C++ Solution with Detailed Comments

```cpp
// Include all standard C++ libraries.
#include <bits/stdc++.h>

// Use the standard namespace to avoid writing std:: repeatedly.
using namespace std;

// Overload output operator for pairs.
// This helper is not essential for this problem, but is included in the template.
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    // Print the first and second elements separated by a space.
    return out << x.first << ' ' << x.second;
}

// Overload input operator for pairs.
// This helper is also part of the template.
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    // Read the first and second elements.
    return in >> x.first >> x.second;
}

// Overload input operator for vectors.
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    // Read every element of the vector.
    for(auto& x: a) {
        in >> x;
    }

    // Return the input stream.
    return in;
};

// Overload output operator for vectors.
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    // Print every element followed by a space.
    for(auto x: a) {
        out << x << ' ';
    }

    // Return the output stream.
    return out;
};

// Side length of the large triangle.
int n;

// cnt[i] stores how many triangles remain of internal kind i.
// Internal kind i has exactly i black edges.
// Input kinds 1..4 are stored as indices 0..3.
array<int, 4> cnt;

// Read input.
void read() {
    // Read side length.
    cin >> n;

    // Read counts of the 4 triangle kinds.
    for(int i = 0; i < 4; i++) {
        cin >> cnt[i];
    }
}

// v_edge[r][p] stores the shared edge between cells p and p+1 in row r.
// It is the right side of cell p and the left side of cell p+1.
array<array<int, 16>, 8> v_edge{};

// h_edge[r][k] stores the shared edge between row r and row r+1.
// It is the bottom edge of upward cell (r, 2k)
// and the top edge of downward cell (r+1, 2k+1).
array<array<int, 16>, 8> h_edge{};

// Memoization table.
// The key packs the current position, active edge mask, and remaining counts.
// The value is the number of completions from that state.
unordered_map<int64_t, int64_t> memo;

// Compute the active edge mask before placing cell (r, p).
// Only these previously assigned edges can affect the future.
int64_t active_mask(int r, int p) {
    // Resulting bitmask.
    int64_t mask = 0;

    // Current bit position inside the mask.
    int bit = 0;

    // If r >= 2, there may be horizontal edges from the previous row
    // that have not yet been consumed by downward triangles in this row.
    if(r >= 2) {
        // Previous-row horizontal edges h_edge[r-1][k] are consumed by
        // downward cells in current row.
        //
        // At position p, edges with k < p/2 have already been consumed.
        // Edges from p/2 to r-2 may still matter.
        for(int k = p / 2; k <= r - 2; k++) {
            // Store the color bit of this edge.
            mask |= (int64_t)(h_edge[r - 1][k] & 1) << bit++;
        }
    }

    // If p > 0, the current cell has a left neighbor.
    // Its left edge is already fixed by v_edge[r][p-1].
    if(p > 0) {
        // Store that left-edge color in the active mask.
        mask |= (int64_t)(v_edge[r][p - 1] & 1) << bit++;
    }

    // Count how many upward triangles in the current row have already been placed.
    //
    // Upward cells are at even positions 0, 2, 4, ...
    // Before position p, the number of built upward cells is (p + 1) / 2.
    int built = (p + 1) / 2;

    // Their bottom edges will be needed by the next row.
    for(int k = 0; k < built; k++) {
        // Store h_edge[r][k] in the active mask.
        mask |= (int64_t)(h_edge[r][k] & 1) << bit++;
    }

    // Return the compact active boundary.
    return mask;
}

// Encode the DP state into one integer key.
int64_t encode_key(int r, int p, int64_t mask) {
    // Convert the cell position into a linear-ish index.
    //
    // Rows before r contain:
    // 1 + 3 + 5 + ... + (2r - 3) = (r - 1)^2 cells.
    // So the current cell index is (r - 1)^2 + p.
    int64_t key = (r - 1) * (r - 1) + p;

    // Add the active edge mask.
    // 16 bits is more than enough for N <= 5.
    key = (key << 16) | mask;

    // Add the remaining counts.
    for(int i = 0; i < 4; i++) {
        // Each count is at most 25, so 5 bits are enough.
        key = (key << 5) | cnt[i];
    }

    // Return the packed key.
    return key;
}

// Recursive DP.
// Returns the number of valid completions starting before cell (r, p).
int64_t rec(int r, int p) {
    // If r is past the last row, all cells have been placed successfully.
    if(r > n) {
        return 1;
    }

    // Compute the memoization key using position, active boundary, and counts.
    int64_t key = encode_key(r, p, active_mask(r, p));

    // Check whether this state has already been solved.
    auto it = memo.find(key);

    // If yes, return the stored answer.
    if(it != memo.end()) {
        return it->second;
    }

    // Compute the next cell in row-major order.
    int next_r = r;
    int next_p = p + 1;

    // Row r has 2r - 1 cells, with last index 2r - 2.
    // If next_p reaches 2r - 1, move to the next row.
    if(next_p >= 2 * r - 1) {
        next_r = r + 1;
        next_p = 0;
    }

    // Even positions are upward triangles, odd positions are downward triangles.
    bool is_up = (p % 2 == 0);

    // The current cell's left edge is fixed if there is a left neighbor.
    // Otherwise there is no constraint, represented by -1.
    int fixed_left = (p > 0) ? v_edge[r][p - 1] : -1;

    // For a downward triangle, its top edge is fixed by the previous row.
    // For an upward triangle, there is no top-edge constraint.
    int fixed_top = is_up ? -1 : h_edge[r - 1][(p - 1) / 2];

    // Accumulate the number of valid completions.
    int64_t total = 0;

    // Try each internal kind.
    // kind means the number of black edges: 0, 1, 2, or 3.
    for(int kind = 0; kind < 4; kind++) {
        // Cannot use a kind if none remain.
        if(cnt[kind] == 0) {
            continue;
        }

        // This kind requires exactly kind black edges.
        int num_black = kind;

        // Enumerate all possible assignments of colors to the 3 sides.
        // Bit 0 = left edge.
        // Bit 1 = right edge.
        // Bit 2 = third edge, which is bottom for upward and top for downward.
        for(int mask = 0; mask < 8; mask++) {
            // Only orientations with the correct number of black edges are valid.
            if(__builtin_popcount(mask) != num_black) {
                continue;
            }

            // Extract the left-edge color.
            int e_left = (mask >> 0) & 1;

            // Extract the right-edge color.
            int e_right = (mask >> 1) & 1;

            // Extract the third-edge color.
            int e_third = (mask >> 2) & 1;

            // If the left edge was already fixed, it must match.
            if(fixed_left != -1 && fixed_left != e_left) {
                continue;
            }

            // If this is a downward triangle, the top edge was already fixed.
            // The third edge is the top edge for downward triangles.
            if(!is_up && fixed_top != e_third) {
                continue;
            }

            // Store the right edge for the next cell in the same row.
            v_edge[r][p] = e_right;

            // If this is an upward triangle, its third edge is its bottom edge.
            // It will be used by a downward triangle in the next row.
            if(is_up) {
                h_edge[r][p / 2] = e_third;
            }

            // Use one triangle of this kind.
            cnt[kind]--;

            // Recurse to the next cell and add all completions.
            total += rec(next_r, next_p);

            // Backtrack: restore the count.
            cnt[kind]++;
        }
    }

    // Store the computed answer in the memoization table.
    memo[key] = total;

    // Return the number of completions.
    return total;
}

// Solve one test case.
void solve() {
    /*
        The big triangle has n rows.

        Row r contains 2r - 1 unit triangles.
        Positions are 0-based.

        Even positions are upward-facing triangles.
        Odd positions are downward-facing triangles.

        The DP fills the board in row-major order.

        At each step, we try all possible triangle kinds and orientations,
        respecting already-fixed shared edges.

        Memoization uses:
        - current cell
        - remaining counts
        - active boundary mask
    */

    // Clear memoization table.
    memo.clear();

    // Start before the first cell: row 1, position 0.
    cout << rec(1, 0) << '\n';
}

// Main function.
int main() {
    // Speed up C++ input/output.
    ios_base::sync_with_stdio(false);

    // Untie cin from cout for faster input.
    cin.tie(nullptr);

    // Number of test cases.
    // The official input has one test case.
    int T = 1;

    // If multiple test cases were needed, this line could be enabled.
    // cin >> T;

    // Process each test case.
    for(int test = 1; test <= T; test++) {
        // Read input.
        read();

        // Solve and print answer.
        solve();
    }

    // Successful program termination.
    return 0;
}
```

---

## 4. Python Solution with Detailed Comments

```python
import sys
from functools import lru_cache


def solve():
    # Read all integers from standard input.
    data = list(map(int, sys.stdin.read().split()))

    # First integer is the side length of the large triangle.
    n = data[0]

    # Next four integers are counts of triangle kinds 1..4.
    #
    # Internally:
    # cnt[0] = number of triangles with 0 black edges
    # cnt[1] = number of triangles with 1 black edge
    # cnt[2] = number of triangles with 2 black edges
    # cnt[3] = number of triangles with 3 black edges
    cnt = data[1:5]

    # v_edge[r][p] stores the shared edge between cells p and p+1 in row r.
    #
    # N <= 5, so small fixed arrays are enough.
    v_edge = [[0] * 16 for _ in range(8)]

    # h_edge[r][k] stores the shared edge between row r and row r+1.
    #
    # It is the bottom side of upward cell (r, 2k)
    # and the top side of downward cell (r+1, 2k+1).
    h_edge = [[0] * 16 for _ in range(8)]

    # Memoization dictionary.
    #
    # Key:
    #   (r, p, active_mask, cnt0, cnt1, cnt2, cnt3)
    #
    # Value:
    #   number of valid completions from this state.
    memo = {}

    def active_mask(r, p):
        """
        Compute a compact bitmask of already-written edges that can still affect
        future placements.

        Edges that will never be read again are intentionally ignored.
        """

        # Result bitmask.
        mask = 0

        # Current bit position.
        bit = 0

        # Horizontal edges from the previous row that have not yet been consumed.
        #
        # h_edge[r-1][k] is consumed by downward cells in row r.
        # At position p, edges with index smaller than p//2 have already been used.
        if r >= 2:
            for k in range(p // 2, r - 1):
                # Store this edge color as one bit.
                mask |= (h_edge[r - 1][k] & 1) << bit
                bit += 1

        # The immediate left edge of the current cell, if there is a left neighbor.
        if p > 0:
            mask |= (v_edge[r][p - 1] & 1) << bit
            bit += 1

        # Bottom edges of upward cells already placed in the current row.
        #
        # These edges will be read by downward cells in the next row.
        built = (p + 1) // 2

        for k in range(built):
            mask |= (h_edge[r][k] & 1) << bit
            bit += 1

        # Return the active boundary.
        return mask

    def rec(r, p):
        """
        Return the number of valid ways to complete the board starting before
        cell (r, p).
        """

        # If we are past the final row, the whole triangle was filled correctly.
        if r > n:
            return 1

        # Current active boundary.
        mask = active_mask(r, p)

        # Memoization key includes the current position, active boundary,
        # and remaining triangle counts.
        key = (r, p, mask, cnt[0], cnt[1], cnt[2], cnt[3])

        # Reuse the result if this state was already solved.
        if key in memo:
            return memo[key]

        # Determine the next cell in row-major order.
        next_r = r
        next_p = p + 1

        # Row r contains 2r - 1 cells.
        # If we passed the last cell, move to row r+1.
        if next_p >= 2 * r - 1:
            next_r = r + 1
            next_p = 0

        # Even positions are upward-facing triangles.
        is_up = (p % 2 == 0)

        # The left edge is fixed by the previous cell in the same row,
        # unless this is the first cell of the row.
        fixed_left = v_edge[r][p - 1] if p > 0 else -1

        # For downward triangles, the top edge is fixed by the row above.
        # For upward triangles, there is no top-edge constraint.
        if is_up:
            fixed_top = -1
        else:
            fixed_top = h_edge[r - 1][(p - 1) // 2]

        # Total number of completions from this state.
        total = 0

        # Try every triangle kind.
        for kind in range(4):
            # Skip if no triangle of this kind remains.
            if cnt[kind] == 0:
                continue

            # Internal kind equals the required number of black edges.
            num_black = kind

            # Enumerate all 3-bit edge color assignments.
            #
            # bit 0: left edge
            # bit 1: right edge
            # bit 2: third edge
            #
            # For upward triangles, third = bottom.
            # For downward triangles, third = top.
            for orient_mask in range(8):
                # This orientation is only valid if it has the required
                # number of black edges.
                if orient_mask.bit_count() != num_black:
                    continue

                # Extract edge colors.
                e_left = (orient_mask >> 0) & 1
                e_right = (orient_mask >> 1) & 1
                e_third = (orient_mask >> 2) & 1

                # Match the already-fixed left edge, if any.
                if fixed_left != -1 and fixed_left != e_left:
                    continue

                # For downward triangles, the third edge is the top edge,
                # which must match the fixed edge from the row above.
                if not is_up and fixed_top != e_third:
                    continue

                # Write the right edge for the next cell in this row.
                v_edge[r][p] = e_right

                # If upward, write the bottom edge for the next row.
                if is_up:
                    h_edge[r][p // 2] = e_third

                # Use one triangle of this kind.
                cnt[kind] -= 1

                # Count all completions after this placement.
                total += rec(next_r, next_p)

                # Backtrack.
                cnt[kind] += 1

        # Store answer for this DP state.
        memo[key] = total

        # Return computed result.
        return total

    # Start at the first row and first position.
    print(rec(1, 0))


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

---

## 5. Compressed Editorial

Process the triangular board row by row. Row `r` has `2r - 1` cells; even positions are upward triangles and odd positions are downward triangles.

Represent edge colors as bits. Internally, kind `k` means a triangle with exactly `k` black edges, for `k = 0..3`. For each placement, enumerate all 3-bit masks with exactly `k` set bits, corresponding to all rotations of that triangle kind.

Maintain shared edges:

- `v_edge[r][p]`: edge between adjacent cells in the same row
- `h_edge[r][k]`: edge between row `r` and row `r + 1`

When placing cell `(r, p)`, its left edge may already be fixed by `v_edge[r][p-1]`. If it is downward-facing, its top edge is fixed by `h_edge[r-1][(p-1)/2]`.

Use DFS with memoization. The full edge history is unnecessary; the future only depends on:

- current position
- remaining counts of each kind
- a small active mask of already-created edges that future cells will still read

Try every valid kind and orientation, write outgoing edges, recurse, then backtrack.

Base case: after row `N`, return `1`.

This works easily for `N ≤ 5` because there are at most 25 cells and only `O(N)` active edge bits.