1. Concise Problem Statement  
Given M axis-aligned wall segments in the plane, each defined by its two endpoints. No two walls overlap except possibly at endpoints. Walls are added one by one in the given order. Find the index (1-based) of the first wall whose addition creates a closed loop (i.e., completes a cycle of connected segments enclosing some area). If no cycle ever forms, output 0.

2. Detailed Editorial  

Overview  
We have an undirected graph whose vertices are the wall endpoints (points with integer coordinates) and whose edges are the walls. We add edges in sequence and must detect the first moment an added edge connects two vertices that were already connected by previous edges—this exactly corresponds to forming a cycle in the graph, hence enclosing some region.

Key Observations  
- A cycle in a graph corresponds to adding an edge between two vertices that are already in the same connected component.  
- We do not need to compute actual enclosed areas or perform any planar embedding; mere connectivity in the graph is sufficient for detection.

Data Structures  
- Disjoint-Set Union (DSU), also known as Union-Find, supports two operations in (amortized) nearly O(1) time:  
  • find(u): returns the representative (root) of u’s set.  
  • union(u,v): merges the sets containing u and v.  
- A hash map that assigns each distinct endpoint (x,y) a unique integer ID on the fly.

Algorithm  
1. Initialize an empty DSU of size up to 2·M (each wall contributes at most 2 new endpoints).  
2. Initialize an empty map `mp` from point→ID.  
3. For each wall i from 1 to M:  
   a. Read its endpoints (x1,y1),(x2,y2).  
   b. If a point is not yet in `mp`, assign it a new ID.  
   c. Let u = ID(x1,y1), v = ID(x2,y2).  
   d. If find(u) == find(v), output i and terminate (cycle formed).  
   e. Else, union(u,v) and continue.  
4. If we finish all M walls without detecting a cycle, output 0.

Complexities  
- Each DSU operation is inverse-Ackermann, effectively constant.  
- Mapping points uses a hash (or balanced map) giving O(log N) or O(1) average per lookup.  
- Total time is O(M α(M) + M log M) or O(M α(M)).  
- Memory is O(M).

3. C++ Solution

```cpp
#include <bits/stdc++.h>
// #include <coding_library/graph/dsu.hpp>

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;
};

class DSU {
  public:
    int n;
    vector<int> par;
    vector<int> sz;

    DSU(int _n = 0) { init(_n); }

    void init(int _n) {
        n = _n;
        par.assign(n + 1, 0);
        sz.assign(n + 1, 0);
        for(int i = 0; i <= n; i++) {
            par[i] = i;
            sz[i] = 1;
        }
    }

    int root(int u) { return par[u] = ((u == par[u]) ? u : root(par[u])); }
    bool connected(int x, int y) { return root(x) == root(y); }

    int unite(int x, int y) {
        x = root(x), y = root(y);
        if(x == y) {
            return x;
        }
        if(sz[x] > sz[y]) {
            swap(x, y);
        }
        par[x] = y;
        sz[y] += sz[x];
        return y;
    }
};

int m;

void read() { cin >> m; }

void solve() {
    // The walls form a planar graph whose vertices are the endpoints; a region
    // becomes enclosed exactly when adding a wall creates the first cycle.
    // Maintain a DSU over endpoints (each distinct coordinate pair gets an id
    // via get_id). For each wall, if its two endpoints are already in the same
    // component the wall closes a cycle, so print its 1-based index; otherwise
    // union the components. If no wall ever closes a cycle, print 0.

    DSU dsu(2 * m);
    map<pair<int, int>, int> mp;

    function<int(int, int)> get_id = [&](int x, int y) {
        if(mp.count({x, y})) {
            return mp[{x, y}];
        }
        return mp[{x, y}] = mp.size();
    };

    for(int i = 0; i < m; i++) {
        int x, y, a, b;
        cin >> x >> y >> a >> b;
        if(dsu.connected(get_id(x, y), get_id(a, b))) {
            cout << i + 1 << '\n';
            return;
        }

        dsu.unite(get_id(x, y), get_id(a, b));
    }

    cout << 0 << '\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 with Detailed Comments  

```python
import sys
sys.setrecursionlimit(10**7)

class DSU:
    def __init__(self, n):
        # parent[i] = parent of i; size[i] = size of tree rooted at i
        self.parent = list(range(n))
        self.size = [1] * n

    def find(self, x):
        # Path compression
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, a, b):
        # Union by size; return True if merged, False if already same set
        ra = self.find(a)
        rb = self.find(b)
        if ra == rb:
            return False
        if self.size[ra] > self.size[rb]:
            ra, rb = rb, ra
        # attach smaller tree ra under rb
        self.parent[ra] = rb
        self.size[rb] += self.size[ra]
        return True

def main():
    data = sys.stdin.read().split()
    it = iter(data)
    M = int(next(it))

    # We'll assign IDs 0.. up to 2*M−1 for endpoints
    dsu = DSU(2 * M)
    point_id = {}    # map (x,y) -> unique id
    next_id = 0

    for i in range(1, M + 1):
        x1 = int(next(it)); y1 = int(next(it))
        x2 = int(next(it)); y2 = int(next(it))
        # Assign/get ID for (x1,y1)
        if (x1,y1) not in point_id:
            point_id[(x1,y1)] = next_id
            next_id += 1
        u = point_id[(x1,y1)]
        # Assign/get ID for (x2,y2)
        if (x2,y2) not in point_id:
            point_id[(x2,y2)] = next_id
            next_id += 1
        v = point_id[(x2,y2)]

        # If they are in the same DSU set, we found the first cycle
        if not dsu.union(u, v):
            print(i)
            return

    # No cycle ever formed
    print(0)

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

5. Compressed Editorial  
Use a Union-Find structure over the wall endpoints. Map each point to a unique ID on demand. As you add each wall (edge), check if its two endpoints are already connected in the DSU; if yes, report the current index (first cycle formed), otherwise merge their sets. If no cycle is detected after all M walls, output 0. This runs in O(M α(M)) time and O(M) memory.