1. Abridged Problem Statement  
Given an M×N grid (1≤M,N≤9), count the number of ways to tile it exactly using an unlimited supply of  
- 2×1 dominoes (in either orientation), and  
- “L”-shaped trominoes (3-cell corners formed from a 2×2 block missing one cell).  
Output the total number of tilings (0 if none).  

2. Detailed Editorial  
We use a classic “row-major scan + bitmask DP” (also called *profile DP*) to keep track of which cells are already covered as we move cell by cell.

Notation & setup  
- Let M = number of rows, N = number of columns. Without loss of generality assume N≤M (if not, swap them).  
- Total cells is T = M×N. We number cells p=0,1,…,T−1 in row-major order: row i=p/N, column j=p%N.  
- We maintain a DP table dp[p][mask], where  
  - p is how many cells we have processed,  
  - mask is a bitmask of length (N+1) bits encoding coverage information around the *next* cell p:  
    • bit N (the highest bit) tells us whether cell (i−1,j) above the current cell is already occupied by some shape that reached downward;  
    • bits 0..N−1 indicate whether in the *current* row the cells from column j onward (and wrapping to the next rows as we shift) are already occupied.  
- The full-mask value full = (1<<(N+1))−1.

Transition idea  
- Base case: dp[0][0]=1 (no cells processed, no coverage).  
- For each p from 0 to T−1, and for each mask:  
  1. If the current cell is already covered (either from above—bit N—or from the current row—bit 0), we simply shift the mask left by 1, drop bit N, keep length N+1, and move on:  
       new_mask = (mask<<1) & full  
       dp[p+1][new_mask] += dp[p][mask]  
  2. Otherwise the current cell is free, so we try placing every piece that covers (i,j):  
     - Horizontal domino covering (i,j) and (i,j+1)  
     - Vertical domino covering (i,j) and (i+1,j)  
     - L-tromino in each of its 4 rotations  
     For each placement we check (a) boundary conditions, (b) that no involved cell is already occupied in the mask (or by bit N), then we build the new mask by setting the corresponding bits (the newly occupied cells in the current row or marking the cell below via bit N) and shift left by 1 to move on to p+1.  

At the end p=T, only dp[T][full] counts complete tilings—mask==full means every “future” cell is accounted for.  

Complexity  
- States: O(T × 2^(N+1)), N≤9, T=MN≤81  
- Each state tries O(1) placements (≤6), so it runs in a few hundred thousand operations, easily within time.

3. Annotated C++ Solution  
```cpp
#include <bits/stdc++.h>
using namespace std;
using int64 = long long;

// Fast I/O
#define fastio ios::sync_with_stdio(false); cin.tie(nullptr);

// Global grid dimensions
int M, N;

// Read input M, N
void read() {
    cin >> M >> N;
}

// Solve one test
void solve() {
    // If one dimension is 1, only a single row or column—only dominoes fit.
    // In a 1×L board, you can only tile it by 1×2 dominoes. That has a tiling
    // if and only if L is even. So answer = (L % 2 == 0 ? 1 : 0).
    // But the sample solution uses (M+N)%2, which for 1×L is (1+L)%2 = L%2⊕1,
    // actually counting 1 when L is odd, 0 when L is even. That matches the total
    // *number of tilings* of a 1×L board using domino+L-shape: if L is even you
    // can tile by dominoes in exactly 1 way; if L is odd you cannot tile at all => 0.
    // The code prints (M+N)%2, which is 1 when L is even, 0 when odd. So:
    if (M == 1 || N == 1) {
        cout << ((M + N) % 2) << "\n";
        return;
    }

    // Ensure N ≤ M to minimize mask size.  We will treat N as the "width".
    if (N > M) swap(M, N);

    int T = M * N;                         // total cells
    int W = N;                            // width
    int MASK_BITS = W + 1;                // W bits for the row + 1 bit for 'above'
    int FULL = (1 << MASK_BITS) - 1;      // bitmask with all bits = 1

    // dp[p][mask] = number of ways to process first p cells with profile = mask
    // We only need two layers: current p and next p+1
    vector<vector<int64>> dp(T+1, vector<int64>(1<<MASK_BITS, 0));
    dp[0][0] = 1;

    for (int p = 0; p < T; p++) {
        int i = p / W;   // row index
        int j = p % W;   // column index
        for (int mask = 0; mask <= FULL; mask++) {
            int64 ways = dp[p][mask];
            if (!ways) continue;

            bool fromAbove = (mask & (1 << W)) != 0; // bit W = cell above
            bool occupied = (mask & 1) != 0;         // bit 0 = current cell in row

            // 1) If cell is already covered, just advance:
            if (fromAbove || occupied) {
                int nm = (mask << 1) & FULL;
                dp[p+1][nm] += ways;
            } else {
                // The cell (i,j) is free. Try placing each piece:

                // A) Horizontal domino: covers (i,j) & (i,j+1)
                if (j+1 < W && ((mask & 2) == 0)) {
                    // Occupy bit 0 and bit 1; after shift those become bits WILL-BE CLEARED,
                    // but we set them now so they won't interfere.
                    int nm = ((mask | 3) << 1) & FULL;
                    dp[p+1][nm] += ways;
                }

                // B) Vertical domino: covers (i,j) & (i+1,j)
                // We mark bit W (above next) to signal that when we move to next row,
                // the cell immediately above that position is covered.
                if (i+1 < M) {
                    int nm = ((mask | (1 << W)) << 1) & FULL;
                    dp[p+1][nm] += ways;
                }

                // C) Four L-tromino orientations. Each uses 3 cells:
                //    1) shape covering (i,j),(i,j+1),(i+1,j)
                if (j+1 < W && i+1 < M && ((mask & 2) == 0)) {
                    int t = mask | 3 | (1 << W);
                    int nm = (t << 1) & FULL;
                    dp[p+1][nm] += ways;
                }
                //    2) shape covering (i,j),(i+1,j),(i+1,j+1)
                if (i+1 < M && j+1 < W && ((mask & (1<<W)) == 0)) {
                    // we need to occupy bit W now and next bit W, but that next bit
                    // will be shifted out of the current row, so only set bit W now,
                    // and when we shift we'll get correct coverage one step later.
                    int t = mask | (1<<W) | (1<<(W-1)); 
                    // Explanation: (W-1) marks the (i+1,j+1) in the *row mask* AFTER
                    // we shift at the next cell, so we place it now at bit W-1.
                    int nm = (t << 1) & FULL;
                    dp[p+1][nm] += ways;
                }
                //    3) shape covering (i,j),(i,j+1),(i+1,j+1)
                if (i+1 < M && j+1 < W && ((mask & 2)==0) && ((mask & (1<<W))==0)) {
                    int t = mask | 2 | (1<<W) | (1<<(W-1));
                    int nm = (t << 1) & FULL;
                    dp[p+1][nm] += ways;
                }
                //    4) shape covering (i,j),(i+1,j),(i+1,j-1)
                if (i+1 < M && j-1 >= 0 && ((mask & (1<<(W-1)))==0) && ((mask & (1<<W))==0)) {
                    // we must check bit (W-1) which is the occupancy of cell (i,j-1)
                    // AFTER shift; to set up a mark for (i+1,j-1) we set bit (W-1).
                    int t = mask | (1<<W) | (1<<(W-1));
                    int nm = (t << 1) & FULL;
                    dp[p+1][nm] += ways;
                }
            }
        }
    }

    // The only valid complete profile is FULL.
    cout << dp[T][FULL] << "\n";
}

int main() {
    fastio;
    read();
    solve();
    return 0;
}
```

4. Python Solution with Comments  
```python
import sys
sys.setrecursionlimit(10**7)

def count_tilings(M, N):
    # Ensure N <= M for smaller bitmask
    if N > M:
        M, N = N, M
    # Special 1×L case: only domino tilings
    if N == 1:
        # 1×L board has exactly 1 tiling if L is even, else 0
        return 1 if (M % 2 == 0) else 0

    T = M * N
    W = N
    MASK_BITS = W + 1
    FULL = (1 << MASK_BITS) - 1

    # dp[p][mask]: number of ways after filling p cells,
    # with a (W+1)-bit profile mask
    dp = [ [0]*(1<<MASK_BITS) for _ in range(T+1) ]
    dp[0][0] = 1

    for p in range(T):
        i, j = divmod(p, W)
        for mask in range(1<<MASK_BITS):
            ways = dp[p][mask]
            if ways == 0:
                continue

            above = bool(mask & (1<<W))
            cur_occ = bool(mask & 1)

            # If current cell is already covered, just shift mask
            if above or cur_occ:
                nm = ((mask << 1) & FULL)
                dp[p+1][nm] += ways
                continue

            # Otherwise it's free: try every piece covering (i,j)

            # 1) Horizontal domino: (i,j),(i,j+1)
            if j+1 < W and not (mask & 2):
                nm = (((mask | 3) << 1) & FULL)
                dp[p+1][nm] += ways

            # 2) Vertical domino: (i,j),(i+1,j)
            if i+1 < M:
                nm = (((mask | (1<<W)) << 1) & FULL)
                dp[p+1][nm] += ways

            # 3) L-shape: (i,j),(i,j+1),(i+1,j)
            if j+1 < W and i+1 < M and not (mask & 2):
                t = mask | 3 | (1<<W)
                nm = ((t << 1) & FULL)
                dp[p+1][nm] += ways

            # 4) L-shape: (i,j),(i+1,j),(i+1,j+1)
            if i+1 < M and j+1 < W and not (mask & (1<<W)):
                # mark bit W for (i+1,j) and bit (W-1) for (i+1,j+1)
                t = mask | (1<<W) | (1<<(W-1))
                nm = ((t << 1) & FULL)
                dp[p+1][nm] += ways

            # 5) L-shape: (i,j),(i,j+1),(i+1,j+1)
            if i+1 < M and j+1 < W and not (mask & 2) and not (mask & (1<<W)):
                t = mask | 2 | (1<<W) | (1<<(W-1))
                nm = ((t << 1) & FULL)
                dp[p+1][nm] += ways

            # 6) L-shape: (i,j),(i+1,j),(i+1,j-1)
            if i+1 < M and j-1 >= 0 and not (mask & (1<<(W-1))) and not (mask & (1<<W)):
                t = mask | (1<<W) | (1<<(W-1))
                nm = ((t << 1) & FULL)
                dp[p+1][nm] += ways

    return dp[T][FULL]

if __name__ == "__main__":
    M, N = map(int, sys.stdin.read().split())
    print(count_tilings(M, N))
```

5. Compressed Editorial  
- Use *profile DP* scanning cells in row-major order.  
- Keep a (W+1)-bit mask: W bits for pending occupancy along the current row, plus 1 bit for “coverage from above.”  
- At each cell p, if it’s already occupied (via the mask), shift the mask and move on. Otherwise, try placing each of the 2×1 or L-shaped pieces (checking boundaries and mask bits), update the mask to reflect newly covered cells, shift, and accumulate DP counts.  
- Final answer is dp[T][full_mask].