1. Abridged Problem Statement
   Given N wolves and M sheep, each represented by a line segment strictly in the half‐plane y ≥ 1. You stand at the origin (0,0) and can fire rays (shots) in any direction. A shot kills every animal whose segment the ray intersects. Find the minimum number of shots required to kill all wolves without intersecting any sheep. If it is impossible, output “No solution.”

2. Detailed Editorial
   a) Geometry → Angle Intervals
   - Every segment above the x-axis can be “seen” from the origin under a continuous range of angles. For a segment with endpoints (x₁,y₁), (x₂,y₂), compute α = atan2(y₁,x₁), β = atan2(y₂,x₂), and let its angular interval be [min(α,β), max(α,β)].
   - A shot at angle θ kills exactly those segments whose intervals contain θ.
   - Since every endpoint has y ≥ 1, all angles are strictly positive (between 0 and π). This means we never need to worry about wrap-around: the angles form a simple range and we can sort them linearly starting from 0.

   b) Sheep as Forbidden Intervals
   - Any angle inside a sheep’s interval is forbidden. Merge all sheep intervals into a sorted list of non‐overlapping forbidden intervals.

   c) Wolves as Target Intervals
   - Each wolf also gives an interval [w₁,w₂]. We must choose angles θ₁ < θ₂ < … such that each wolf-interval contains at least one chosen θ, and no θ lies inside any forbidden sheep interval.

   d) Greedy Interval Stabbing with Obstacles
   - Sort wolves by their right endpoint ascending. Maintain “last” as the angle of the most recent shot (initialized to –∞).
   - For each wolf in order:
     • If its left endpoint ≤ last, the wolf is already killed by a previous shot—skip.
     • Otherwise, we must place a new shot within [w_left, w_right], strictly greater than last, but not inside a sheep interval.
     • Advance a pointer through the merged sheep intervals to the first one whose right end is at least w_right. If its left end is below w_right, we can only shoot as late as (s_left – ε); otherwise we can shoot at w_right.
     • If this computed shooting angle < w_left, it’s impossible → “No solution.” Else place the shot there and increment the count.

   e) Correctness & Complexity
   - This is the classic greedy for minimum interval “stabbing” points, extended to forbid certain sub‐ranges. Shooting as far right as possible always maximizes coverage of future wolves.
   - Merging sheep intervals takes O(M log M). Sorting wolves takes O(N log N). The one‐pass scan with two pointers over wolves and sheep is O(N + M).

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 double eps = 1e-9;

int n, m;
vector<pair<double, double>> wolves, sheep;

void read() {
    cin >> n >> m;
    wolves.resize(n);
    sheep.resize(m);
    for(auto& [alpha, beta]: wolves) {
        int x1, y1, x2, y2;
        cin >> x1 >> y1 >> x2 >> y2;
        alpha = atan2((double)y1, (double)x1);
        beta = atan2((double)y2, (double)x2);
        if(alpha > beta) {
            swap(alpha, beta);
        }
    }
    for(auto& [alpha, beta]: sheep) {
        int x1, y1, x2, y2;
        cin >> x1 >> y1 >> x2 >> y2;
        alpha = atan2((double)y1, (double)x1);
        beta = atan2((double)y2, (double)x2);
        if(alpha > beta) {
            swap(alpha, beta);
        }
    }
}

void solve() {
    // Convert 2D problem to 1D angle ranges and use greedy approach:
    // Union overlapping sheep ranges, sort wolves by right endpoint,
    // greedily place each wolf as far right as possible without hitting sheep.
    //
    // Note that in this problem we have Y >= 1, meaning that all angles will be
    // positive. It also means that we don't actually need to look at all
    // rotations to run the greedy for, and we can simply sort by atan2 angle
    // starting from 0.0.

    if(!sheep.empty()) {
        sort(sheep.begin(), sheep.end());
        vector<pair<double, double>> merged;
        auto current = sheep[0];

        for(int i = 1; i < (int)sheep.size(); i++) {
            if(sheep[i].first <= current.second + eps) {
                current.second = max(current.second, sheep[i].second);
            } else {
                merged.push_back(current);
                current = sheep[i];
            }
        }
        merged.push_back(current);
        sheep = merged;
    }

    sort(wolves.begin(), wolves.end(), [](const auto& a, const auto& b) {
        return a.second < b.second;
    });

    int pos_sheep = 0;
    double last = -1e9;
    int count = 0;

    for(auto& wolf: wolves) {
        if(wolf.first < last + eps) {
            continue;
        }

        while(pos_sheep < (int)sheep.size() &&
              wolf.second > sheep[pos_sheep].second) {
            pos_sheep++;
        }

        double rightmost = wolf.second;
        if(pos_sheep < (int)sheep.size()) {
            rightmost = min(rightmost, sheep[pos_sheep].first - eps);
        }

        if(rightmost < wolf.first - eps) {
            cout << "No solution\n";
            return;
        }

        last = rightmost;
        count++;
    }

    cout << count << "\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
import math

def main():
    data = sys.stdin.read().split()
    if not data:
        return
    it = iter(data)
    n, m = int(next(it)), int(next(it))

    wolves = []
    sheep = []

    # Helper to read one segment and turn into an angular interval
    def read_interval():
        x1, y1, x2, y2 = map(int, (next(it), next(it), next(it), next(it)))
        a = math.atan2(y1, x1)
        b = math.atan2(y2, x2)
        return (min(a, b), max(a, b))

    for _ in range(n):
        wolves.append(read_interval())
    for _ in range(m):
        sheep.append(read_interval())

    # Merge sheep intervals into non-overlapping forbidden zones
    sheep.sort()
    merged = []
    for interval in sheep:
        if not merged or interval[0] > merged[-1][1] + 1e-9:
            merged.append(list(interval))
        else:
            # Overlap: extend the end
            merged[-1][1] = max(merged[-1][1], interval[1])
    sheep = merged

    # Sort wolves by right endpoint for greedy stabbing
    wolves.sort(key=lambda x: x[1])

    last_shot = -1e18
    shots = 0
    pos_sheep = 0
    S = len(sheep)

    for wl, wr in wolves:
        # Already covered?
        if wl <= last_shot + 1e-9:
            continue

        # Skip sheep intervals that end before this wolf’s right edge
        while pos_sheep < S and sheep[pos_sheep][1] < wr:
            pos_sheep += 1

        # Best shot is at wr, but must avoid the next sheep interval
        shoot = wr
        if pos_sheep < S:
            s_left = sheep[pos_sheep][0]
            shoot = min(shoot, s_left - 1e-9)

        # If shoot < wl, no feasible shot
        if shoot < wl - 1e-9:
            print("No solution")
            return

        # Place shot
        last_shot = shoot
        shots += 1

    print(shots)

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

5. Compressed Editorial
   • Convert each segment to an angular interval [α,β] via atan2.
   • Merge sheep intervals into forbidden zones.
   • Sort wolves by β, then greedily choose each shot at the rightmost feasible angle ≤ β but > previous shot and outside any sheep interval.
   • If at any step no valid angle remains inside [α,β], answer “No solution,” else print shot count.
