1. **Abridged problem statement**

You are given up to 50 spheres in 3D space (center and radius), a starting point of a laser ray, and another point that defines its initial direction. The ray:

- travels in straight lines,
- reflects perfectly on sphere surfaces (angle of incidence = angle of reflection),
- may hit the same sphere multiple times,
- is considered to have reflected even if it only grazes a sphere tangentially.

The ray starts strictly outside all spheres. Simulate the ray and output the sequence of indices of the spheres it hits, in order. If it reflects more than 10 times, output only the first 10 indices followed by `etc.`.



2. **Detailed editorial**

### 2.1 Geometry model

We work in 3D with:

- Points and vectors represented as `(x, y, z)`.
- A ray defined by:
  \[
  P(t) = O + t\cdot D, \quad t \ge 0
  \]
  where:
  - `O` is the current origin (point where the ray currently starts),
  - `D` is the current direction vector (we normalize it for numeric stability).

Each sphere `i` is defined by:

- center `C_i = (x_i, y_i, z_i)`
- radius `r_i`

We need to:

1. Find which sphere is hit next by the current ray.
2. Compute the hit point.
3. Compute the reflection direction at that point.
4. Move to that point and repeat for up to 11 hits (we simulate at most 11 to know if >10 happened).

Important subtlety: after hitting a sphere, the ray could immediately intersect the same sphere again if we start the new ray from inside or numerically on the surface. To avoid counting the same surface as an immediate second hit, we skip the last-hit sphere in the next search.

### 2.2 Ray–sphere intersection

Sphere equation:

\[
|P - C|^2 = r^2
\]

Ray: \(P(t) = O + tD\).

Let \(V = O - C\). Plug in:

\[
|O + tD - C|^2 = r^2 \\
|V + tD|^2 = r^2 \\
(V + tD)\cdot (V + tD) = r^2 \\
V\cdot V + 2t (V\cdot D) + t^2 (D\cdot D) = r^2
\]

Rearrange to:

\[
a t^2 + b t + c = 0
\]
where
- \(a = D\cdot D\)
- \(b = 2 (V\cdot D)\)
- \(c = V\cdot V - r^2\)

Discriminant:

\[
\Delta = b^2 - 4ac
\]

Cases:

- If \(\Delta < 0\): no intersection.
- If \(\Delta = 0\): tangent (one intersection); we must treat this as a valid hit.
- If \(\Delta > 0\): two intersection parameters:

  \[
  t_1 = \frac{-b - \sqrt{\Delta}}{2a}, \quad t_2 = \frac{-b + \sqrt{\Delta}}{2a}
  \]

We want the **smallest positive** `t` (the nearest point in front of the ray origin):

- If `t1 > EPS` (EPS ~ 1e-6), return `t1`.
- Else if `t2 > EPS`, return `t2`.
- Else, no valid positive intersection.

We also clamp the discriminant with `max(0, disc)` before `sqrt` to guard against tiny negative values due to floating-point error.

**Tangents:** If ray just touches a sphere, `Δ` is ~0, so both `t1` and `t2` are equal (within numeric precision) and one of them will be > EPS; we treat that as a hit.

### 2.3 Choosing the next sphere

For each step:

- Iterate over all spheres `i` (0..n-1).
- Skip `i == last_hit` (the sphere just hit in the previous step), to avoid immediately re-intersecting the same surface due to numerical issues.
- Compute `t = ray_sphere_intersect(origin, dir, sphere[i])`.
- Only consider `t > 0`.
- Track the minimum `t` and its sphere index.

If no sphere has a positive intersection parameter, the ray escapes, and we stop.

### 2.4 Computing reflection

At the hit point:

- Hit point:
  \[
  P = O + tD
  \]
- Surface normal at that point:

  \[
  N = \frac{P - C}{\|P - C\|}
  \]

  (A unit vector pointing from the sphere center to the hit point.)

For incoming direction `D` (normalized), reflection direction `R` is given by:

\[
R = D - 2(D\cdot N) N
\]

We normalize `R` again to reduce error accumulation.

Then we set:

- `origin = P`
- `dir = R`

and repeat.

### 2.5 Simulation details

We simulate up to 11 bounces:

- We want to output up to the first 10 hits. 
- We simulate one extra step (11th hit) to check if there were more than 10. This is equivalent in effect to: simulate while `step < 11`.

For each found hit, store `(sphere_index + 1)` (1-based indexing) in `hits`.

After the loop:

- If `len(hits) <= 10`, print all hit indices separated by spaces.
- If `len(hits) > 10`, print the first 10 indices then ` etc.`.

### 2.6 Edge cases and numeric issues

- Starting point is guaranteed to be outside any sphere, so initial intersections always have `t > 0` if intersection exists.
- Using an `EPS` of `1e-6` avoids counting an intersection behind the origin (`t ≈ 0` or very small negative) as a forward hit.
- We use double precision (`double`) everywhere.
- We normalize direction vectors whenever they are created/updated.
- We do not special-case tangents; the quadratic math already treats them as hits.

### 2.7 Complexity

Each step we:

- Intersect the ray with all `n` spheres: `O(n)` operations.
- We simulate at most 11 steps, so complexity is `O(11n) = O(n)` with `n ≤ 50`, easily within time limits.

Memory usage is negligible.



3. **Commented C++ solution**

```cpp
#include <bits/stdc++.h>          // Includes almost all standard headers; convenient for contests

using namespace std;

// Overload operator<< for pair<T1,T2> to print "first second"
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

// Overload operator>> for pair<T1,T2> to read "first second"
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

// Overload operator>> for vector<T> to read all elements in order
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;                 // Uses existing operator>> for element type
    }
    return in;
};

// Overload operator<< for vector<T> to print all elements separated by space
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

// Small epsilon for floating point comparisons
const double EPS = 1e-6;

// 3D vector / point structure
struct Vec3 {
    double x, y, z;

    // Constructor with default values (0,0,0)
    Vec3(double x = 0, double y = 0, double z = 0) : x(x), y(y), z(z) {}

    // Vector addition
    Vec3 operator+(const Vec3& o) const {
        return Vec3(x + o.x, y + o.y, z + o.z);
    }

    // Vector subtraction
    Vec3 operator-(const Vec3& o) const {
        return Vec3(x - o.x, y - o.y, z - o.z);
    }

    // Scalar multiplication: v * k
    Vec3 operator*(double k) const {
        return Vec3(x * k, y * k, z * k);
    }

    // Dot product with another vector
    double dot(const Vec3& o) const {
        return x * o.x + y * o.y + z * o.z;
    }

    // Euclidean norm (length) of the vector
    double norm() const {
        return sqrt(dot(*this)); // sqrt(x^2 + y^2 + z^2)
    }

    // Return normalized (unit length) vector
    Vec3 normalize() const {
        double n = norm();
        return Vec3(x / n, y / n, z / n);
    }
};

// Sphere: center and radius
struct Sphere {
    Vec3 center;
    double r;
};

// Global variables
int n;                      // number of spheres
vector<Sphere> spheres;     // list of spheres
Vec3 origin, dir;           // current ray origin and direction (unit vector)

// Compute intersection of ray with sphere
double ray_sphere_intersect(Vec3 O, Vec3 D, const Sphere& s) {
    // Ray: P(t) = O + t*D, t >= 0
    // Sphere: |P - C|^2 = r^2
    // Let V = O - C, then we solve quadratic:
    //   a*t^2 + b*t + c = 0
    // Where:
    //   a = D.D
    //   b = 2 * (V.D)
    //   c = V.V - r^2
    // We return the smallest t > EPS, or -1 if no valid intersection.

    Vec3 V = O - s.center;       // Vector from sphere center to ray origin
    double a = D.dot(D);         // Should be 1 if D is normalized, but we compute generally
    double b = 2 * V.dot(D);
    double c = V.dot(V) - s.r * s.r;

    double disc = b * b - 4 * a * c;  // Discriminant

    // If discriminant < 0 (with small tolerance), no real intersection
    if(disc < -EPS) {
        return -1;
    }

    // Guard against tiny negative due to precision; clamp to zero
    disc = sqrt(max(0.0, disc));

    // Two candidate solutions
    double t1 = (-b - disc) / (2 * a);
    double t2 = (-b + disc) / (2 * a);

    // We want the smallest positive t (greater than EPS)
    if(t1 > EPS) {
        return t1;
    }
    if(t2 > EPS) {
        return t2;
    }

    // No positive intersection
    return -1;
}

// Compute reflection of direction D about normal N
Vec3 reflect(Vec3 D, Vec3 N) {
    // Reflection formula:
    // R = D - 2(D·N)N
    return D - N * (2 * D.dot(N));
}

// Read input
void read() {
    cin >> n;                     // number of spheres
    spheres.resize(n);

    // Read each sphere's center and radius
    for(int i = 0; i < n; i++) {
        cin >> spheres[i].center.x >> spheres[i].center.y >>
            spheres[i].center.z >> spheres[i].r;
    }

    // Read two points:
    // (x1,y1,z1) = starting point of ray
    // (x2,y2,z2) = another point on ray, defines direction
    double x1, y1, z1, x2, y2, z2;
    cin >> x1 >> y1 >> z1 >> x2 >> y2 >> z2;

    // Set origin
    origin = Vec3(x1, y1, z1);

    // Direction = (second point - first point), then normalize
    dir = Vec3(x2 - x1, y2 - y1, z2 - z1).normalize();
}

// Core simulation and output
void solve() {
    // We simulate the path of the ray as it reflects from spheres.

    vector<int> hits;        // list of indices of spheres hit (1-based)
    int last_hit = -1;       // index of most recently hit sphere (0-based), -1 = none

    // Simulate up to 11 interactions
    for(int step = 0; step < 11; step++) {
        double best_t = 1e18;  // large initial "infinite" distance
        int best_sphere = -1;  // index of sphere with closest intersection

        // For every sphere, find intersection with current ray
        for(int i = 0; i < n; i++) {
            // Skip the last hit sphere to avoid immediately re-hitting
            if(i == last_hit) {
                continue;
            }

            double t = ray_sphere_intersect(origin, dir, spheres[i]);

            // We want smallest positive t
            if(t > 0 && t < best_t) {
                best_t = t;
                best_sphere = i;
            }
        }

        // If no sphere is intersected, stop simulation
        if(best_sphere == -1) {
            break;
        }

        // Record the sphere number (1-based index)
        hits.push_back(best_sphere + 1);

        // Update last_hit so we skip it next iteration
        last_hit = best_sphere;

        // Compute intersection point P = origin + dir * best_t
        Vec3 P = origin + dir * best_t;

        // Compute normal at hit point, pointing outward from sphere center
        Vec3 N = (P - spheres[best_sphere].center).normalize();

        // Compute reflection direction and normalize for stability
        dir = reflect(dir, N).normalize();

        // New ray origin is the hit point
        origin = P;
    }

    // We only print up to the first 10 hits
    int output_count = min((int)hits.size(), 10);

    // Print hit sequence separated by spaces
    for(int i = 0; i < output_count; i++) {
        if(i > 0) {
            cout << " ";
        }
        cout << hits[i];
    }

    // If more than 10 hits, append " etc."
    if(hits.size() > 10) {
        cout << " etc.";
    }

    cout << "\n";
}

int main() {
    ios_base::sync_with_stdio(false); // Fast IO
    cin.tie(nullptr);                 // Untie cin from cout

    int T = 1;                        // Number of test cases; fixed to 1 here
    // cin >> T;                      // Could support multiple tests if needed
    for(int test = 1; test <= T; test++) {
        read();                      // Read one test case
        // cout << "Case #" << test << ": ";
        solve();                     // Solve and output result
    }

    return 0;                        // Successful termination
}
```

4. **Python solution with detailed comments**

```python
import sys
import math

EPS = 1e-6

class Vec3:
    """Simple 3D vector class with basic operations."""
    __slots__ = ("x", "y", "z")

    def __init__(self, x=0.0, y=0.0, z=0.0):
        self.x = float(x)
        self.y = float(y)
        self.z = float(z)

    def __add__(self, other):
        """Vector addition: self + other"""
        return Vec3(self.x + other.x, self.y + other.y, self.z + other.z)

    def __sub__(self, other):
        """Vector subtraction: self - other"""
        return Vec3(self.x - other.x, self.y - other.y, self.z - other.z)

    def __mul__(self, k):
        """Scalar multiplication: self * k"""
        return Vec3(self.x * k, self.y * k, self.z * k)

    # For k * v, Python will call __rmul__
    __rmul__ = __mul__

    def dot(self, other):
        """Dot product with another Vec3."""
        return self.x * other.x + self.y * other.y + self.z * other.z

    def norm(self):
        """Euclidean length of the vector."""
        return math.sqrt(self.dot(self))

    def normalize(self):
        """Return a normalized (unit length) copy of the vector."""
        n = self.norm()
        # Problem guarantees we won't have zero-length directions
        return Vec3(self.x / n, self.y / n, self.z / n)


class Sphere:
    """Sphere with center (Vec3) and radius r."""
    __slots__ = ("center", "r")

    def __init__(self, center, r):
        self.center = center
        self.r = float(r)


def ray_sphere_intersect(O, D, sphere):
    """
    Compute intersection of ray P(t) = O + t*D with sphere.
    Return smallest t > EPS, or -1 if no such intersection.
    """
    # Vector from sphere center to ray origin
    V = O - sphere.center

    # Quadratic coefficients: a*t^2 + b*t + c = 0
    a = D.dot(D)              # usually 1.0 if D is normalized
    b = 2.0 * V.dot(D)
    c = V.dot(V) - sphere.r * sphere.r

    # Discriminant
    disc = b * b - 4.0 * a * c

    # No real roots: discriminant negative beyond tolerance
    if disc < -EPS:
        return -1.0

    # Clamp small negative values to 0 before sqrt for numerical safety
    disc = math.sqrt(max(0.0, disc))

    # Two candidate intersection parameters
    t1 = (-b - disc) / (2.0 * a)
    t2 = (-b + disc) / (2.0 * a)

    # We need smallest positive t (t > EPS)
    if t1 > EPS:
        return t1
    if t2 > EPS:
        return t2

    # Both are behind or very close to origin -> no valid intersection
    return -1.0


def reflect(D, N):
    """
    Reflect direction D around normal N.
    Reflection formula: R = D - 2(D·N)N
    Both D and N are Vec3.
    """
    k = 2.0 * D.dot(N)
    return D - (k * N)


def solve_one_case(data_iter):
    """
    Read a single test case from iterator of tokens and print result.
    The problem as given uses a single test per file.
    """
    # Read number of spheres
    try:
        n = int(next(data_iter))
    except StopIteration:
        return False  # no more data

    spheres = []

    # Read n spheres: xi yi zi ri
    for _ in range(n):
        x = float(next(data_iter))
        y = float(next(data_iter))
        z = float(next(data_iter))
        r = float(next(data_iter))
        spheres.append(Sphere(Vec3(x, y, z), r))

    # Last line: 6 real numbers for two points
    x1 = float(next(data_iter))
    y1 = float(next(data_iter))
    z1 = float(next(data_iter))
    x2 = float(next(data_iter))
    y2 = float(next(data_iter))
    z2 = float(next(data_iter))

    # Ray origin
    origin = Vec3(x1, y1, z1)

    # Direction from first point to second, normalized
    dir_vec = Vec3(x2 - x1, y2 - y1, z2 - z1).normalize()

    hits = []          # list of sphere indices (1-based) hit in order
    last_hit = -1      # index (0-based) of last hit sphere; -1 if none

    # Simulate up to 11 hits (one extra to detect "more than 10")
    for step in range(11):
        best_t = float('inf')  # nearest intersection distance along ray
        best_idx = -1          # index of sphere giving that intersection

        # Check intersection with each sphere
        for i, s in enumerate(spheres):
            # Skip last-hit sphere to avoid immediate re-hitting the same one
            if i == last_hit:
                continue

            t = ray_sphere_intersect(origin, dir_vec, s)

            # We only care about positive t; choose the smallest
            if t > 0.0 and t < best_t:
                best_t = t
                best_idx = i

        # If we didn't find any intersection, the ray escapes
        if best_idx == -1:
            break

        # Record 1-based index of this sphere
        hits.append(best_idx + 1)

        # Update last hit
        last_hit = best_idx

        # Compute hit point P = origin + dir_vec * best_t
        P = origin + dir_vec * best_t

        # Compute outward normal at the hit point and normalize
        N = (P - spheres[best_idx].center).normalize()

        # Compute reflected direction and normalize to unit length
        dir_vec = reflect(dir_vec, N).normalize()

        # New ray origin is the hit point
        origin = P

    # Output logic: print first up to 10 hits, separated by spaces
    out_hits = hits[:10]
    if out_hits:
        sys.stdout.write(" ".join(str(x) for x in out_hits))
    # If more than 10 hits, append " etc."
    if len(hits) > 10:
        if out_hits:
            sys.stdout.write(" ")
        sys.stdout.write("etc.")
    sys.stdout.write("\n")

    return True


def main():
    # Read all input as tokens (whitespace-separated)
    data = sys.stdin.read().strip().split()
    if not data:
        return

    it = iter(data)

    # Problem specifies a single test case, but we make it generic:
    # if multiple cases were concatenated, this would process them all.
    while True:
        if not solve_one_case(it):
            break


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

5. **Compressed editorial**

- Represent the ray as \(P(t) = O + tD\) with normalized direction `D`.
- For each sphere, solve intersection by plugging the ray into \(|P - C|^2 = r^2\). This yields a quadratic in `t`. Compute discriminant:
  - If `< 0`: no intersection.
  - Otherwise compute `t1`, `t2`; choose smallest `t > EPS`.
- For each step, among all spheres except the last one hit, pick the sphere with minimal positive `t`. If none exists, stop.
- For a hit:
  - Record its 1-based index.
  - Compute hit point `P = O + tD`.
  - Compute normal `N = (P - C).normalize()`.
  - Reflect direction: `D = (D - 2(D·N)N).normalize()`.
  - Set new origin `O = P`.
- Simulate up to 11 bounces (for up to 10 outputs plus detection of “more than 10”). Print first 10 indices; if more hits exist, append `etc.`.