1. Abridged Problem Statement  
Given an N×N grid initially all white, perform M repaint operations. Each operation specifies a rectangle by its corners (x₁,y₁) and (x₂,y₂) and a color C (‘w’ for white, ‘b’ for black). After all operations, output the total number of white cells.

2. Detailed Editorial

Overview  
We must support up to N=1000 and M=5000 rectangle-paint operations on an N×N grid and then count white cells. A naïve per-cell update in each rectangle (O(M·N²)) would be too slow. Instead we exploit bit-parallelism: represent each row as a bitset of length N, where bit j=1 means “white” and 0 means “black.” Then:

- To paint columns y₁…y₂ white in row i, we OR that row’s bitset with a mask having 1s in [y₁,y₂].
- To paint them black, we AND with the bitwise complement of that mask.

A single row update is O(N/word_size)≈O(N/64). Applying this for all rows x₁…x₂ makes a rectangle update O((x₂–x₁+1)·N/64). In the worst case M=5000 and each rectangle covers almost all rows, total cost is M·(N/64)·N≈5000·(1000/64)≈78 million 64-bit operations. This passes comfortably in 1.25 s in C++.  

Further optimization (used in the provided C++ code) is a √-decomposition over rows: group rows into blocks of size B≈√N. For each block, maintain two lazy bitsets: one for pending ORs and one for pending ANDs. When a rectangle fully covers an entire block, we update the block’s lazy bitsets in O(N/64) time instead of touching each row. When it partially covers a block, we first “push” (apply) that block’s lazies to its rows, then do per-row updates. Finally, we push all lazies and count bits.

Step-by-step  
1. Initialize each row’s bitset to all 1s (white).  
2. Compute block size sq = ⌊√N⌋+1, number of blocks ≈ ⌈N/sq⌉.  
3. For each operation (x₁,y₁)-(x₂,y₂), normalize coordinates so x₁≤x₂, y₁≤y₂ and build mask = ((1<<(y₂−y₁+1))−1) << y₁.  
4. Identify block indices r₁ = x₁/sq, r₂ = x₂/sq.  
   - If r₁=r₂, push that block’s lazy into rows and update rows x₁…x₂ directly with OR or AND.  
   - Else: push lazies for r₁ and r₂, update the partial rows at the ends; for each fully covered block between r₁+1 and r₂−1, update its lazy OR/AND bitsets.  
5. After all operations, push all lazies to rows, then sum the counts of 1-bits in each row.

3. Provided C++ Solution with Detailed Comments

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

// Overload << for pair to ease debugging (not used in final solution)
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Overload >> for pair (not essential here)
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Overload >> for vector
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

// Overload << for vector
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

const int B = 1024;           // Max grid width (bitset size)

int n, m;                     // Grid size and number of operations
int sq;                       // Block size ≈ sqrt(n)
vector<bitset<B>> grid;       // grid[i] stores row i’s bits
vector<bitset<B>> set_1_lazy; // Lazy OR masks per block
vector<bitset<B>> set_0_lazy; // Lazy AND masks per block
bitset<B> full_m;             // mask with first n bits = 1

// Apply pending lazy updates for block 'bucket' to its rows
void apply_lazy(
    vector<bitset<B>>& grid,
    vector<bitset<B>>& set_1_lazy,
    vector<bitset<B>>& set_0_lazy,
    int bucket,
    int n,
    int sq
) {
    int start = bucket * sq;                    // first row in this block
    int end = min(n, (bucket + 1) * sq);        // one past last row
    for(int i = start; i < end; i++) {
        // OR with pending whites, then AND with complement of pending blacks
        grid[i] |= set_1_lazy[bucket];
        grid[i] &= ~set_0_lazy[bucket];
    }
    // Clear lazies once applied
    set_1_lazy[bucket].reset();
    set_0_lazy[bucket].reset();
}

// (Optional) debugging printer to show current grid state
void print_table(
    vector<bitset<B>>& grid, 
    vector<bitset<B>>& set_1_lazy,
    vector<bitset<B>>& set_0_lazy, 
    int n, 
    int sq
) {
    // First push all lazies
    for(int bucket = 0; bucket * sq < n; bucket++) {
        apply_lazy(grid, set_1_lazy, set_0_lazy, bucket, n, sq);
    }
    // Print rows as W/B characters
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < n; j++) {
            cout << (grid[i][j] ? 'W' : 'B');
        }
        cout << '\n';
    }
    cout << '\n';
}

// Read input
void read() {
    cin >> n >> m;
}

void solve() {
    sq = sqrt(n) + 1;                      // choose block size
    grid.assign(n, bitset<B>());           // n empty bitsets
    set_1_lazy.assign(sq, bitset<B>());    // lazies per block
    set_0_lazy.assign(sq, bitset<B>());

    // Build a bitset with first n bits = 1
    full_m.reset();
    for(int i = 0; i < n; i++) {
        full_m.set(i);
    }
    // Initialize all rows to white
    for(int i = 0; i < n; i++) {
        grid[i] = full_m;
    }

    // Process each repaint operation
    while(m--) {
        int x1, y1, x2, y2;
        string c;
        cin >> x1 >> y1 >> x2 >> y2 >> c; // read corners and color
        // Convert to 0-based
        x1--; y1--; x2--; y2--;
        if(x1 > x2) swap(x1, x2);
        if(y1 > y2) swap(y1, y2);

        bool color = (c == "w"); // true=white, false=black
        // Build mask with bits y1..y2 set
        bitset<B> mask = (full_m >> (n - (y2 - y1 + 1))) << y1;

        int r1 = x1 / sq;  // block index of top row
        int r2 = x2 / sq;  // block index of bottom row

        if(r1 == r2) {
            // Entirely within one block: push its lazy
            apply_lazy(grid, set_1_lazy, set_0_lazy, r1, n, sq);
            // Update rows x1..x2 directly
            for(int i = x1; i <= x2; i++) {
                if(color) grid[i] |= mask;
                else       grid[i] &= ~mask;
            }
        } else {
            // Left partial block
            apply_lazy(grid, set_1_lazy, set_0_lazy, r1, n, sq);
            for(int i = x1; i < (r1 + 1) * sq && i <= x2; i++) {
                if(color) grid[i] |= mask;
                else       grid[i] &= ~mask;
            }
            // Right partial block
            apply_lazy(grid, set_1_lazy, set_0_lazy, r2, n, sq);
            for(int i = r2 * sq; i <= x2; i++) {
                if(color) grid[i] |= mask;
                else       grid[i] &= ~mask;
            }
            // Fully covered blocks in between
            for(int b = r1 + 1; b < r2; b++) {
                if(color) {
                    // Mark to OR mask later, clear any pending AND on those bits
                    set_1_lazy[b] |= mask;
                    set_0_lazy[b] &= ~mask;
                } else {
                    // Mark to AND-mask-out later, clear any pending OR on those bits
                    set_0_lazy[b] |= mask;
                    set_1_lazy[b] &= ~mask;
                }
            }
        }

        // (Optional) debug print
        // print_table(grid, set_1_lazy, set_0_lazy, n, sq);
    }

    // Push remaining lazies to actual rows
    for(int bucket = 0; bucket * sq < n; bucket++) {
        apply_lazy(grid, set_1_lazy, set_0_lazy, bucket, n, sq);
    }
    // Count white bits in each row
    long long ans = 0;
    for(int i = 0; i < n; i++) {
        ans += grid[i].count();
    }
    cout << ans << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    read();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments

```python
import sys
input = sys.stdin.readline

def main():
    n, m = map(int, input().split())
    # rows[i] is a Python int whose binary representation stores row i
    # Bit j=1 means white, 0 means black.
    rows = [(1 << n) - 1 for _ in range(n)]  # initialize all n bits to 1

    for _ in range(m):
        x1, y1, x2, y2, c = input().split()
        x1 = int(x1)-1; y1 = int(y1)-1
        x2 = int(x2)-1; y2 = int(y2)-1
        # Ensure x1<=x2 and y1<=y2
        if x1 > x2:
            x1, x2 = x2, x1
        if y1 > y2:
            y1, y2 = y2, y1

        length = y2 - y1 + 1
        # mask with length 1s, then shift to y1
        mask = ((1 << length) - 1) << y1

        if c == 'w':
            # paint white: OR with mask
            for i in range(x1, x2+1):
                rows[i] |= mask
        else:
            # paint black: AND with complement of mask
            comp = ~mask
            for i in range(x1, x2+1):
                rows[i] &= comp

    # Count total white bits
    total = 0
    for r in rows:
        # built-in bit_count() in Python 3.8+
        total += r.bit_count()
    print(total)

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

5. Compressed Editorial  
- Represent each row as an N-bit bitmask.  
- Build for any rectangle the column-mask `( (1<<(y2−y1+1))−1 )<<y1`.  
- For each row in [x1…x2], do `row |= mask` for white or `row &= ~mask` for black.  
- Bitwise operations on 64-bit (or Python int) chunks make this fast.  
- Optionally speed up further via √-decomposition and lazy masks on row-blocks.