## 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. 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;
array<int, 4> cnt;

void read() {
    cin >> n;
    for(int i = 0; i < 4; i++) {
        cin >> cnt[i];
    }
}

array<array<int, 16>, 8> v_edge{};
array<array<int, 16>, 8> h_edge{};
unordered_map<int64_t, int64_t> memo;

int64_t active_mask(int r, int p) {
    int64_t mask = 0;
    int bit = 0;
    if(r >= 2) {
        for(int k = p / 2; k <= r - 2; k++) {
            mask |= (int64_t)(h_edge[r - 1][k] & 1) << bit++;
        }
    }
    if(p > 0) {
        mask |= (int64_t)(v_edge[r][p - 1] & 1) << bit++;
    }
    int built = (p + 1) / 2;
    for(int k = 0; k < built; k++) {
        mask |= (int64_t)(h_edge[r][k] & 1) << bit++;
    }
    return mask;
}

int64_t encode_key(int r, int p, int64_t mask) {
    int64_t key = (r - 1) * (r - 1) + p;
    key = (key << 16) | mask;
    for(int i = 0; i < 4; i++) {
        key = (key << 5) | cnt[i];
    }
    return key;
}

int64_t rec(int r, int p) {
    if(r > n) {
        return 1;
    }

    int64_t key = encode_key(r, p, active_mask(r, p));
    auto it = memo.find(key);
    if(it != memo.end()) {
        return it->second;
    }

    int next_r = r, next_p = p + 1;
    if(next_p >= 2 * r - 1) {
        next_r = r + 1;
        next_p = 0;
    }

    bool is_up = (p % 2 == 0);
    int fixed_left = (p > 0) ? v_edge[r][p - 1] : -1;
    int fixed_top = is_up ? -1 : h_edge[r - 1][(p - 1) / 2];

    int64_t total = 0;
    for(int kind = 0; kind < 4; kind++) {
        if(cnt[kind] == 0) {
            continue;
        }

        int num_black = kind;
        for(int mask = 0; mask < 8; mask++) {
            if(__builtin_popcount(mask) != num_black) {
                continue;
            }

            int e_left = (mask >> 0) & 1;
            int e_right = (mask >> 1) & 1;
            int e_third = (mask >> 2) & 1;

            if(fixed_left != -1 && fixed_left != e_left) {
                continue;
            }
            if(!is_up && fixed_top != e_third) {
                continue;
            }

            v_edge[r][p] = e_right;
            if(is_up) {
                h_edge[r][p / 2] = e_third;
            }

            cnt[kind]--;
            total += rec(next_r, next_p);
            cnt[kind]++;
        }
    }

    memo[key] = total;
    return total;
}

void solve() {
    // The big triangle has n rows; row r (1..n) contains 2r-1 unit cells at
    // positions 0..2r-2, with upward triangles at even positions and downward
    // triangles at odd positions. Edges shared between cells:
    //
    //   - v_edge[r][p] is the vertical edge between cells p and p+1 of row r
    //     (the right side of cell p, the left side of cell p+1);
    //
    //   - h_edge[r][k] is the horizontal edge between row r and row r+1, with
    //     k indexing the k-th such edge from the left; it equals the bottom
    //     side of the upward cell at (r, 2k) and the top side of the downward
    //     cell at (r+1, 2k+1).
    //
    // Kind 1+k has exactly k black edges out of 3 (k = 0..3); placing such a
    // triangle in any orientation means choosing which of (left, right, third)
    // sides receive the k black colors, where the "third" side is the bottom
    // for an upward cell and the top for a downward cell.
    //
    // We DP over cells in row-major order. At the moment we are about to
    // place cell (r, p), the future is determined by the remaining counts
    // cnt[0..3] together with a small "active border" of already-determined
    // edges that future cells will still read:
    //
    //   - h_edge[r-1][k] for k = floor(p/2) .. r-2 — bottoms of row r-1 that
    //     have not yet been consumed by row r's downward cells;
    //
    //   - v_edge[r][p-1] when p > 0 — the left side that the current cell
    //     must match;
    //
    //   - h_edge[r][k] for k = 0 .. ceil(p/2)-1 — bottoms of row r's upward
    //     cells already placed, which will be read by row r+1.
    //
    // That mask has at most n+1 bits (six bits for n=5). Memoizing on (cell
    // index, mask, cnt[0..3]) collapses the search: for each placement we
    // enumerate every (kind, orientation), keep those consistent with the
    // fixed left/top edges, decrement the kind count, and recurse.

    memo.clear();
    cout << rec(1, 0) << '\n';
}

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 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.
