## 1) A concise, abridged problem statement

You are given **N** pizza huts at integer coordinates \((x_i, y_i)\). Hut \(i\) can deliver within Manhattan distance \(w_i\), i.e. it covers all points \((x,y)\) with
\[
|x-x_i| + |y-y_i| \le w_i.
\]
A hut \(A\) **can be closed** if **at least K other huts** can reach \(A\) within their own \(w\) (within 10 minutes).

Output how many huts can be closed and their 1-based indices in increasing order.

Constraints: \(2 \le N \le 10^5\), \(1 \le K \le N-1\), coordinates and \(w\) up to \(10^9\).

---

## 2) Detailed editorial (solution explanation)

### Key geometry transformation (diamond → axis-aligned rectangle)

The set of points within Manhattan distance \(w\) from \((x,y)\) is a **diamond**:
\[
|X-x| + |Y-y| \le w.
\]

Use the classic 45° transform:
\[
u = x + y,\quad v = x - y.
\]

For any point \((X,Y)\), define \((U,V)\) similarly. Then:
\[
|X-x| + |Y-y| \le w \iff
\begin{cases}
|U-u| \le w\\
|V-v| \le w
\end{cases}
\]
So in \((u,v)\)-space, each diamond becomes an **axis-aligned rectangle**:
\[
u \in [u-w,\, u+w],\quad v \in [v-w,\, v+w].
\]

Now the problem becomes:

For each hut center point \((u_i, v_i)\), count how many rectangles (from all huts \(j\)) contain that point. Hut \(i\) is closable iff:
\[
(\text{count containing } (u_i,v_i)) - 1 \ge K
\]
(the “-1” excludes its own rectangle, which always contains its center).

---

### Sweep line over \(u\) + Fenwick (BIT) over \(v\)

We need point-in-rectangle counts for up to \(10^5\) rectangles and \(10^5\) query points.

Standard method:
- Sweep increasing \(u\).
- Maintain which rectangles are currently active (whose \(u\)-interval covers current sweep position).
- For active rectangles, we need to count how many cover a given \(v\).

For a rectangle:
- It becomes active at \(u = u_{\min} = u_i - w_i\)
- It becomes inactive just after \(u_{\max} = u_i + w_i\)

To handle inclusivity cleanly with integers, we use “end at \(u_{\max}+1\)”:
- add event at \(u_i - w_i\): +1 (activate)
- add event at \(u_i + w_i + 1\): -1 (deactivate)

When active, it covers \(v \in [v_i - w_i,\, v_i + w_i]\).

---

### Range add + point query on \(v\)

While sweeping, we need to support:
- Add \(\Delta\) to all \(v\) in an interval \([L, R]\) (for activation/deactivation)
- Query the value at a single \(v_i\)

This can be done with a Fenwick tree using the difference-array trick:
- To add \(\Delta\) to range \([L,R]\):  
  `diff[L] += Δ`, `diff[R+1] -= Δ`
- Point value at \(p\) is prefix sum up to \(p\).

So for each active rectangle, we do:
- update at index(L): +Δ
- update at index(R+1): -Δ

Then for each query point \(v_i\), answer is `prefix_sum(index(v_i))`.

---

### Coordinate compression on \(v\)

\(v\) values can be up to \(10^9\) and negative, so we compress all \(v\)-coordinates we’ll ever address in Fenwick updates/queries:
We need coordinates for:
- \(v_i - w_i\) (range start)
- \(v_i + w_i + 1\) (range end+1 point for diff trick)
- \(v_i\) (query point)

Collect all these \(3N\) values, sort unique, map to 1..M.

---

### Event ordering correctness

We create 3 events per hut \(i\):
1. At \(u_i - w_i\): rectangle update with \(\Delta = +1\)
2. At \(u_i + w_i + 1\): rectangle update with \(\Delta = -1\)
3. At \(u_i\): query for point \((u_i, v_i)\)

When multiple events share the same \(u\):
- We must process **updates before queries** so rectangles that start at \(u\) are counted.
- With the “+1 end” construction, rectangles ending at \(u_{\max}+1\) should be removed **before** queries at that \(u\), which is fine since they no longer cover that \(u\).

Thus sorting by `(u, type)` where `type=0` for update and `type=1` for query works.

Complexity:
- Building events: \(O(N)\)
- Sorting events: \(O(N \log N)\)
- Each event does \(O(\log N)\) Fenwick ops → total \(O(N \log N)\)
- Memory: \(O(N)\)

---

## 3) Provided C++ solution with detailed line-by-line comments

```cpp
#include <bits/stdc++.h>
// #include <coding_library/data_structures/fenwick.hpp>

using namespace std;

// Print a pair as "first second"
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Read a pair from input
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Read an entire vector (assumes it is already sized)
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

// Print a vector with spaces (trailing space included)
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

// Fenwick Tree (Binary Indexed Tree) supporting point updates + prefix sums.
// Also includes find_kth (not used in this problem).
template<class T>
class Fenwick {
  private:
    int sz, log_size;      // sz is internal size; log_size for find_kth
    vector<T> tr;          // fenwick internal array (1-indexed)

  public:
    // Initialize tree for n elements (1..n)
    void init(int n) {
        sz = n + 1;                    // internal limit (a bit oversized)
        log_size = 31 - __builtin_clz(sz); // floor(log2(sz))
        tr.assign(sz + 1, 0);          // allocate and fill with 0
    }

    // Add val to index idx (idx must be >= 1)
    void update(int idx, T val) {
        if(idx <= 0) {                 // defensive check
            assert(false);
            return;
        }
        // Standard Fenwick update loop
        for(; idx <= sz; idx += (idx & -idx)) {
            tr[idx] += val;
        }
    }

    // Prefix sum query: sum of [1..idx]
    T query(int idx) {
        T ans = 0;
        // Standard Fenwick prefix sum loop
        for(; idx >= 1; idx -= (idx & -idx)) {
            ans += tr[idx];
        }
        return ans;
    }

    // Range sum query [l..r] (not used in this solution)
    T query(int l, int r) { return query(r) - query(l - 1); }

    // Find smallest idx such that prefix_sum(idx) >= k (not used here)
    int find_kth(T k) {
        int idx = 0;
        for(int i = log_size; i >= 0; i--) {
            if(idx + (1 << i) < sz && tr[idx + (1 << i)] < k) {
                k -= tr[idx + (1 << i)];
                idx += (1 << i);
            }
        }
        return idx + 1;
    }
};

int n, k;                 // number of huts and threshold K
vector<int64_t> x, y, w;   // input arrays

// Read input
void read() {
    cin >> n >> k;
    x.resize(n);
    y.resize(n);
    w.resize(n);
    for(int i = 0; i < n; i++) {
        cin >> x[i] >> y[i] >> w[i];
    }
}

void solve() {
    // Transform coordinates:
    // u = x + y, v = x - y
    // Each Manhattan ball becomes rectangle:
    // u in [u-w, u+w], v in [v-w, v+w]
    vector<int64_t> tx(n), ty(n);  // tx = u, ty = v
    for(int i = 0; i < n; i++) {
        tx[i] = x[i] + y[i];
        ty[i] = x[i] - y[i];
    }

    // Coordinate compression for all v-values we will touch in Fenwick:
    //   v-w, v+w+1 (for diff trick), and v itself (query point).
    vector<int64_t> all_ys;
    for(int i = 0; i < n; i++) {
        all_ys.push_back(ty[i] - w[i]);      // interval start
        all_ys.push_back(ty[i] + w[i] + 1);  // interval end+1 (diff array)
        all_ys.push_back(ty[i]);             // query point
    }
    sort(all_ys.begin(), all_ys.end());
    all_ys.erase(unique(all_ys.begin(), all_ys.end()), all_ys.end());

    // Lambda that maps a coordinate value to compressed index in [1..M]
    auto compress = [&](int64_t v) -> int {
        return lower_bound(all_ys.begin(), all_ys.end(), v) - all_ys.begin() + 1;
    };

    // Event structure:
    // x_coord: sweep coordinate u
    // type: 0 = rectangle update event, 1 = query event
    // idx: hut index
    // delta: +1 for activate, -1 for deactivate (only for update events)
    struct Event {
        int64_t x_coord;
        int type;
        int idx;
        int delta;
    };

    vector<Event> events;
    events.reserve(3 * n);

    // For each hut i (rectangle):
    // activate at u-w, deactivate at u+w+1, query at u
    for(int i = 0; i < n; i++) {
        events.push_back({tx[i] - w[i], 0, i, 1});       // start rectangle
        events.push_back({tx[i] + w[i] + 1, 0, i, -1});  // end rectangle (exclusive)
        events.push_back({tx[i], 1, i, 0});              // query its center
    }

    // Sort by u, and for ties process updates (type=0) before queries (type=1)
    sort(events.begin(), events.end(), [](const Event& a, const Event& b) {
        if(a.x_coord != b.x_coord) {
            return a.x_coord < b.x_coord;
        }
        return a.type < b.type;
    });

    // Fenwick over compressed v-coordinates.
    // We store a "difference array" in Fenwick so that:
    // range add [L,R] by Δ => add Δ at L, add -Δ at (R+1).
    Fenwick<int> fw;
    fw.init((int)all_ys.size());

    vector<int> cnt(n); // cnt[i] = number of rectangles covering point (u_i,v_i)

    // Sweep through events in increasing u
    for(auto& e: events) {
        if(e.type == 0) {
            // Rectangle update event for hut e.idx with delta e.delta
            // It affects v in [v-w, v+w], implemented as diff:
            // diff[lo] += delta; diff[hi] -= delta, where hi is (v+w+1).
            int lo = compress(ty[e.idx] - w[e.idx]);
            int hi = compress(ty[e.idx] + w[e.idx] + 1);
            fw.update(lo, e.delta);
            fw.update(hi, -e.delta);
        } else {
            // Query event: how many active rectangles cover v_i?
            // That's the prefix sum at coordinate v_i.
            cnt[e.idx] = fw.query(compress(ty[e.idx]));
        }
    }

    // Determine which huts can be closed:
    // cnt[i] includes its own rectangle, so subtract 1.
    vector<int> result;
    for(int i = 0; i < n; i++) {
        if(cnt[i] - 1 >= k) {
            result.push_back(i + 1); // convert to 1-based index
        }
    }

    // Output count and indices (already increasing because i increases)
    cout << result.size() << "\n";
    for(int i = 0; i < (int)result.size(); i++) {
        if(i) {
            cout << " ";
        }
        cout << result[i];
    }
    cout << "\n";
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T; // only one test in this problem
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}
```

---

## 4) Python solution (same approach) with detailed comments

```python
import sys
from bisect import bisect_left

# Fenwick Tree (Binary Indexed Tree) for prefix sums.
class Fenwick:
    def __init__(self, n: int):
        # Use 1-based indexing, size = n
        self.n = n
        self.bit = [0] * (n + 1)

    def add(self, i: int, delta: int) -> None:
        # Adds delta to index i
        while i <= self.n:
            self.bit[i] += delta
            i += i & -i

    def sum(self, i: int) -> int:
        # Returns prefix sum [1..i]
        s = 0
        while i > 0:
            s += self.bit[i]
            i -= i & -i
        return s


def solve() -> None:
    data = sys.stdin.buffer.read().split()
    it = iter(data)
    n = int(next(it))
    k = int(next(it))

    x = [0] * n
    y = [0] * n
    w = [0] * n
    for i in range(n):
        x[i] = int(next(it))
        y[i] = int(next(it))
        w[i] = int(next(it))

    # Transform to (u, v) coordinates
    u = [0] * n
    v = [0] * n
    for i in range(n):
        u[i] = x[i] + y[i]
        v[i] = x[i] - y[i]

    # Collect all v-coordinates needed for compression:
    # v-w (range start), v+w+1 (end+1 for diff), v (query)
    coords = []
    for i in range(n):
        coords.append(v[i] - w[i])
        coords.append(v[i] + w[i] + 1)
        coords.append(v[i])
    coords = sorted(set(coords))

    # Compression function: value -> 1-based index
    def comp(val: int) -> int:
        return bisect_left(coords, val) + 1

    # Build events:
    # type 0 = update, type 1 = query (updates must come before queries at same u)
    events = []
    for i in range(n):
        events.append((u[i] - w[i], 0, i, 1))        # activate rectangle
        events.append((u[i] + w[i] + 1, 0, i, -1))   # deactivate (exclusive end)
        events.append((u[i], 1, i, 0))               # query point

    # Sort by u, then type (0 before 1)
    events.sort()

    fw = Fenwick(len(coords))
    cnt = [0] * n  # number of rectangles covering each point

    # Sweep line over u
    for ux, typ, idx, delta in events:
        if typ == 0:
            # Range add on v in [v-w, v+w] using diff:
            # add delta at L, add -delta at (R+1) which is (v+w+1)
            lo = comp(v[idx] - w[idx])
            hi = comp(v[idx] + w[idx] + 1)
            fw.add(lo, delta)
            fw.add(hi, -delta)
        else:
            # Point query at v[idx]
            cnt[idx] = fw.sum(comp(v[idx]))

    # Hut i is closable if at least K OTHER huts cover it => cnt[i]-1 >= K
    res = [str(i + 1) for i in range(n) if cnt[i] - 1 >= k]

    # Output
    out = []
    out.append(str(len(res)))
    out.append(" ".join(res))
    sys.stdout.write("\n".join(out) + "\n")


if __name__ == "__main__":
    solve()
```

---

## 5) Compressed editorial

Transform each hut \((x,y)\) to \((u,v)=(x+y, x-y)\). The Manhattan ball \(|X-x|+|Y-y|\le w\) becomes rectangle \(u\in[u-w,u+w]\), \(v\in[v-w,v+w]\). For each hut center \((u_i,v_i)\), we need how many rectangles contain it; closable iff count−1 ≥ K.

Do an \(u\)-sweep with events:
- at \(u_i-w_i\): activate rectangle (+1)
- at \(u_i+w_i+1\): deactivate rectangle (−1)
- at \(u_i\): query point

Maintain coverage along \(v\) using a Fenwick tree on compressed \(v\)-coordinates with range-add/point-query via difference trick:
add Δ at \(L=v-w\), add −Δ at \(R+1=v+w+1\); query is prefix sum at \(v_i\).

Sort events by \(u\), updates before queries. Total \(O(N\log N)\), memory \(O(N)\).