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. 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, m;

void read() { cin >> n >> m; }

void solve() {
    // This is a standard problem for DP with a broken profile, although we need
    // to consider some cases.

    // vector<vector<int64_t>> can_cover(1 << m, vector<int64_t>(1 << m,
    // false));

    // can_cover[0][0] = 1;
    // for(int mask = 0; mask < (1 << m); mask++) {
    //     for(int nmask = 0; nmask < (1 << m); nmask++) {
    //         int i = m - 1;
    //         while(i >= 0 && !(mask & (1 << i)) && !(nmask & (1 << i))) {
    //             i--;
    //         }

    //         if(i == -1) {
    //             continue;
    //         }

    //         if((mask & (1 << i)) && (nmask & (1 << i))) {
    //             can_cover[mask][nmask] +=
    //                 can_cover[mask ^ (1 << i)][nmask ^ (1 << i)];
    //         }

    //         if(i > 0 && (mask & (1 << i)) && (mask & (1 << (i - 1)))) {
    //             can_cover[mask][nmask] +=
    //                 can_cover[mask ^ (1 << i) ^ (1 << (i - 1))][nmask];
    //         }

    //         if(i > 0 && (nmask & (1 << i)) && (nmask & (1 << (i - 1)))) {
    //             can_cover[mask][nmask] +=
    //                 can_cover[mask][nmask ^ (1 << i) ^ (1 << (i - 1))];
    //         }

    //         if(i > 0 && (mask & (1 << i)) && (nmask & (1 << (i - 1))) &&
    //            (nmask & (1 << i))) {
    //             can_cover[mask][nmask] +=
    //                 can_cover[mask ^ (1 << i)]
    //                          [nmask ^ (1 << i) ^ (1 << (i - 1))];
    //         }

    //         if(i > 0 && (mask & (1 << (i - 1))) && (nmask & (1 << i)) &&
    //            (nmask & (1 << (i - 1)))) {
    //             can_cover[mask][nmask] +=
    //                 can_cover[mask ^ (1 << (i - 1))]
    //                          [nmask ^ (1 << i) ^ (1 << (i - 1))];
    //         }

    //         if(i > 0 && (nmask & (1 << i)) && (mask & (1 << (i - 1))) &&
    //            (nmask & (1 << i))) {
    //             can_cover[mask][nmask] +=
    //                 can_cover[mask ^ (1 << (i - 1))]
    //                          [nmask ^ (1 << i) ^ (1 << (i - 1))];
    //         }

    //         if(i > 0 && (nmask & (1 << i)) && (mask & (1 << (i - 1))) &&
    //            (mask & (1 << (i - 1)))) {
    //             can_cover[mask][nmask] +=
    //                 can_cover[mask ^ (1 << (i - 1))]
    //                          [nmask ^ (1 << i) ^ (1 << (i - 1))];
    //         }

    //         cout << bitset<3>(mask) << '\n'
    //              << bitset<3>(nmask) << '\n'
    //              << can_cover[mask][nmask] << '\n';
    //         cout << ">>>> " << i << '\n';
    //         cout << '\n';
    //     }
    // }

    // vector<vector<int64_t>> dp(n + 1, vector<int64_t>(1 << m, 0));
    // dp[0][(1 << m) - 1] = 1;
    // for(int i = 1; i <= n; i++) {
    //     for(int mask = 0; mask < (1 << m); mask++) {
    //         for(int nmask = 0; nmask < (1 << m); nmask++) {
    //             dp[i][nmask] +=
    //                 can_cover[mask ^ ((1 << m) - 1)][nmask] * dp[i -
    //                 1][mask];
    //         }
    //     }
    // }

    if(n == 1 || m == 1) {
        cout << (n + m) % 2 << endl;
        return;
    }

    vector<vector<int64_t>> dp(n * m + 1, vector<int64_t>(1 << (m + 1), 0));
    int total_mask = (1 << (m + 1)) - 1;

    dp[0][0] = 1;
    for(int p = 0; p < n * m; p++) {
        for(int mask = 0; mask < (1 << (m + 1)); mask++) {
            if(!dp[p][mask]) {
                continue;
            }

            int i = p / m, j = p % m;
            if((p > m && (mask & (1 << m))) || p <= m) {
                dp[p + 1][(mask << 1) & total_mask] += dp[p][mask];
            }

            //  XXXXXX
            // X..
            if(j > 0 && !(mask & 1)) {
                if((p > m && (mask & (1 << m))) || p <= m) {
                    dp[p + 1][((mask << 1) | 3) & total_mask] += dp[p][mask];
                }
            }

            //  X.XXXX
            // XX.
            if(i > 0 && !(mask & (1 << (m - 1)))) {
                if((p > m && (mask & (1 << m))) || p <= m) {
                    dp[p + 1][((mask << 1) | 1 | (1 << m)) & total_mask] +=
                        dp[p][mask];
                }
            }

            //  X.XXXX
            // X..
            if(i > 0 && j > 0 && !(mask & (1 << (m - 1))) && !(mask & 1)) {
                if((p > m && (mask & (1 << m))) || p <= m) {
                    dp[p + 1][((mask << 1) | 3 | (1 << m)) & total_mask] +=
                        dp[p][mask];
                }
            }

            //  ..XXXX
            // XX.
            if(i > 0 && j > 0 && !(mask & (1 << (m - 1))) &&
               !(mask & (1 << m))) {
                dp[p + 1][((mask << 1) | 1 | (1 << m)) & total_mask] +=
                    dp[p][mask];
            }

            //  .XXXXX
            // X..
            if(i > 0 && j > 0 && !(mask & (1 << m)) && !(mask & 1)) {
                dp[p + 1][((mask << 1) | 3) & total_mask] += dp[p][mask];
            }

            //  X..XXX
            // XX.
            if(i > 0 && j + 1 < m && !(mask & (1 << (m - 1))) &&
               !(mask & (1 << (m - 2)))) {
                if((p > m && (mask & (1 << m))) || p <= m) {
                    dp[p + 1]
                      [((mask << 1) | 1 | (1 << (m - 1)) | (1 << m)) &
                       total_mask] += dp[p][mask];
                }
            }
        }
    }

    cout << dp[n * m][total_mask] << 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();
        // cout << "Case #" << test << ": ";
        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].