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++.  

The provided C++ code goes one step further with a √-decomposition over rows: group rows into buckets of size B≈√N. For each bucket, maintain two lazy bitsets: one for pending ORs (set_1_lazy, "paint white") and one for pending ANDs (set_0_lazy, "paint black"). When a rectangle fully covers an entire bucket, we update that bucket's lazy bitsets in O(N/64) time instead of touching each row. When it partially covers a bucket, we first "push" (apply) that bucket's lazies to its rows, then do per-row updates. Setting one colour clears the conflicting lazy bits for the same columns. Finally, we push all lazies and count bits.

Step-by-step  
1. Initialize each row's bitset to all 1s (white).  
2. Compute bucket size sq = ⌊√N⌋+1, number of buckets ≈ ⌈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 bucket indices r₁ = x₁/sq, r₂ = x₂/sq.  
   - If r₁=r₂, push that bucket'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 bucket 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. 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;
};

const int B = 1024;

int n, m;
int sq;
vector<bitset<B>> grid;
vector<bitset<B>> set_1_lazy, set_0_lazy;
bitset<B> full_m;

void apply_lazy(int bucket) {
    int start = bucket * sq;
    int end = min(n, (bucket + 1) * sq);
    for(int i = start; i < end; i++) {
        grid[i] |= set_1_lazy[bucket];
        grid[i] &= ~set_0_lazy[bucket];
    }

    set_1_lazy[bucket] = bitset<B>(0);
    set_0_lazy[bucket] = bitset<B>(0);
}

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

void solve() {
    // Sqrt decomposition over rows, each row stored as a bitset of columns.
    // Rows are grouped into buckets of size sq = sqrt(n); each bucket carries
    // two lazy column masks set_1_lazy / set_0_lazy describing pending "set to
    // white" / "set to black" operations that apply uniformly to every row in
    // the bucket. A repaint of rows [x1, x2], columns [y1, y2] touches the two
    // boundary buckets directly (apply_lazy flushes them first, then we edit
    // the affected rows by OR-ing / AND-ing the column mask) and updates only
    // the lazy masks of fully covered interior buckets in O(1). Setting one
    // colour clears the conflicting lazy bits for the same columns. At the end
    // all buckets are flushed and we count the white (set) bits.

    sq = sqrt(n) + 1;
    grid.assign(n, bitset<B>());
    set_1_lazy.assign(sq, bitset<B>());
    set_0_lazy.assign(sq, bitset<B>());

    full_m = bitset<B>();
    for(int i = 0; i < n; i++) {
        full_m.set(i);
    }

    for(int i = 0; i < n; i++) {
        grid[i] = full_m;
    }

    while(m--) {
        int x1, y1, x2, y2;
        string c;
        cin >> x1 >> y1 >> x2 >> y2 >> c;
        x1--;
        y1--;
        x2--;
        y2--;

        if(x1 > x2) {
            swap(x1, x2);
        }

        if(y1 > y2) {
            swap(y1, y2);
        }

        bool color = (c == "w");
        bitset<B> mask = (full_m >> (n - (y2 - y1 + 1))) << y1;

        int r1 = x1 / sq, r2 = x2 / sq;

        if(r1 == r2) {
            apply_lazy(r1);
            for(int i = x1; i <= x2; i++) {
                if(color) {
                    grid[i] |= mask;
                } else {
                    grid[i] &= ~mask;
                }
            }
        } else {
            apply_lazy(r1);
            for(int i = x1; i < (r1 + 1) * sq && i <= x2; i++) {
                if(color) {
                    grid[i] |= mask;
                } else {
                    grid[i] &= ~mask;
                }
            }

            apply_lazy(r2);
            for(int i = r2 * sq; i <= x2; i++) {
                if(color) {
                    grid[i] |= mask;
                } else {
                    grid[i] &= ~mask;
                }
            }

            for(int i = r1 + 1; i < r2; i++) {
                if(color) {
                    set_1_lazy[i] |= mask;
                    set_0_lazy[i] &= ~mask;
                } else {
                    set_0_lazy[i] |= mask;
                    set_1_lazy[i] &= ~mask;
                }
            }
        }
    }

    int ans = 0;
    for(int bucket = 0; bucket * sq < n; bucket++) {
        apply_lazy(bucket);
    }

    for(int i = 0; i < n; i++) {
        ans += grid[i].count();
    }

    cout << ans << '\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();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

4. Python Solution

```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.  
- The C++ code speeds up further via √-decomposition with lazy OR/AND masks on row-buckets.
