## 1. Abridged problem statement

A frog must make two projectile jumps.

- It starts on the ground, `ds` before the first vertical wall.
- The first wall has a hole from height `b1` to `t1`.
- The two walls are distance `l` apart.
- The frog lands somewhere between the walls.
- Then it jumps through the second wall’s hole, from height `b2` to `t2`, and lands on the ground `df` after the second wall.

For each jump, the frog may choose a different direction and starting speed, but both speeds must be at most some maximal speed `v`.

Given `b1, t1, b2, t2, l, ds, df, g`, find the minimum possible `v`.

Input contains multiple test cases until EOF. Output the answer with accuracy `1e-4`.

---

## 2. Detailed editorial

We need minimize the maximum of the required speeds for the two jumps.

The key observation is that once we fix the landing point between the two walls, the two jumps become independent.

Let the frog land at distance `p` after the first wall, where:

```text
0 < p < l
```

Then:

- First jump:
  - distance from start to first wall: `ds`
  - distance from first wall to landing point: `p`
  - hole interval: `[b1, t1]`

- Second jump:
  - distance from landing point to second wall: `l - p`
  - distance from second wall to final landing point: `df`
  - hole interval: `[b2, t2]`

So for a fixed `p`, we can compute the minimum required speed for each jump separately, then take the maximum.

The remaining task is to choose the best `p`.

---

### Single-jump analysis

Consider one jump.

The frog starts at ground level, crosses a wall at horizontal distance `d1`, and lands again at ground level after another distance `d2`.

So the total horizontal jump distance is:

```text
R = d1 + d2
```

The trajectory starts and ends at height `0`, so geometrically it is a parabola with roots at `x = 0` and `x = R`.

Therefore we can write it as:

```text
y(x) = a * x * (R - x)
```

The wall is at `x = d1`, and the frog must pass through the hole at some height `h`.

So:

```text
h = y(d1) = a * d1 * d2
```

Thus:

```text
a = h / (d1 * d2)
```

Now compare this with the physical projectile formula.

Projectile motion:

```text
x(t) = vx * t
y(t) = vy * t - g * t^2 / 2
```

Eliminate `t = x / vx`:

```text
y(x) = (vy / vx) * x - g * x^2 / (2 * vx^2)
```

The coefficient of `x^2` is:

```text
-g / (2 * vx^2)
```

In the geometric form:

```text
y(x) = a * R * x - a * x^2
```

the coefficient of `x^2` is `-a`.

Therefore:

```text
a = g / (2 * vx^2)
```

So:

```text
vx^2 = g / (2a)
```

Also:

```text
vy / vx = a * R
```

Therefore:

```text
vy = vx * a * R
```

The total speed squared is:

```text
v^2 = vx^2 + vy^2
```

Substitute:

```text
v^2 = vx^2 * (1 + a^2 * R^2)
```

Using `vx^2 = g / (2a)`:

```text
v^2 = g / (2a) + g * a * R^2 / 2
```

Since:

```text
a = h / (d1 * d2)
```

we get:

```text
v^2 = g * d1 * d2 / (2h)
    + g * h * (d1 + d2)^2 / (2 * d1 * d2)
```

This has the form:

```text
f(h) = A / h + B * h
```

where:

```text
A = g * d1 * d2 / 2
B = g * (d1 + d2)^2 / (2 * d1 * d2)
```

We need choose `h` inside the hole interval `[b, t]` minimizing `f(h)`.

---

### Optimizing the crossing height

The function:

```text
f(h) = A / h + B * h
```

is convex for `h > 0`.

Differentiate:

```text
f'(h) = -A / h^2 + B
```

Set derivative to zero:

```text
-A / h^2 + B = 0
```

So:

```text
h^2 = A / B
```

After simplification:

```text
h = d1 * d2 / (d1 + d2)
```

This is the best height if there were no hole limits.

But the frog must pass through the hole, so the optimal height is:

```text
h = clamp(d1 * d2 / (d1 + d2), b, t)
```

Then the minimum speed squared for this jump is:

```text
A / h + B * h
```

---

### Optimizing landing point `p`

For a fixed landing point `p`:

```text
first_speed_squared =
    jump(ds, p, b1, t1)

second_speed_squared =
    jump(l - p, df, b2, t2)
```

The frog’s maximal required speed squared is:

```text
cost(p) = max(first_speed_squared, second_speed_squared)
```

We need minimize:

```text
cost(p), 0 < p < l
```

This function is continuous and unimodal in this setting. Moving the landing point to the right makes the first jump longer and the second jump shorter, and vice versa. Therefore ternary search can be used.

Finally, the answer is:

```text
sqrt(min cost(p))
```

Because all distances and hole heights are positive, a feasible pair of jumps always exists with sufficiently large speed, so the provided solution never outputs `-1`.

Complexity per test case:

```text
O(iterations)
```

with around `300` ternary-search iterations, effectively constant time.

---

## 3. Provided C++ solution with detailed comments

```cpp
// Include the whole C++ standard library.
#include <bits/stdc++.h>

// Use the standard namespace to avoid writing std:: everywhere.
using namespace std;

// Output operator for pairs.
// This helper is not actually needed by this problem,
// but it is part of the author's general template.
template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    // Print the first and second elements separated by a space.
    return out << x.first << ' ' << x.second;
}

// Input operator for pairs.
// Also unused in this particular solution.
template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    // Read the first and second elements.
    return in >> x.first >> x.second;
}

// Input operator for vectors.
// Unused here, but included in the template.
template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    // Read every element of the vector.
    for(auto& x: a) {
        in >> x;
    }

    // Return the input stream to allow chaining.
    return in;
};

// Output operator for vectors.
// Unused here.
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    // Print every element followed by a space.
    for(auto x: a) {
        out << x << ' ';
    }

    // Return the output stream to allow chaining.
    return out;
};

// Global variables for one test case.
// b1, t1: lower and upper height of first hole.
// b2, t2: lower and upper height of second hole.
// l: distance between walls.
// ds: distance from starting point to first wall.
// df: distance from second wall to final landing point.
// g: gravity.
long double b1, t1, b2, t2, l, ds, df, g;

// Reads one test case.
bool read() {
    // Try to read all eight real numbers.
    cin >> b1 >> t1 >> b2 >> t2 >> l >> ds >> df >> g;

    // If reading failed, return false.
    // This is how the program stops at EOF.
    return !cin.fail();
}

// Solves one test case.
void solve() {
    // Function that computes the minimum required speed squared for one jump.
    //
    // Parameters:
    // d1 = horizontal distance from jump start to wall
    // d2 = horizontal distance from wall to landing point
    // b  = lower side of the hole
    // t  = upper side of the hole
    auto min_v2_jump = [](long double d1, long double d2, long double b,
                          long double t) {
        // From the derivation:
        // v^2(h) = A / h + B * h.
        long double A = g * d1 * d2 / 2.0L;

        // Coefficient B in the formula.
        long double B = g * (d1 + d2) * (d1 + d2) / (2.0L * d1 * d2);

        // The unconstrained best crossing height is:
        // h* = d1 * d2 / (d1 + d2).
        //
        // But the frog must pass through the hole, so clamp h* into [b, t].
        long double h = max(b, min(t, d1 * d2 / (d1 + d2)));

        // Return the minimum possible speed squared for this jump.
        return A / h + B * h;
    };

    // cost(p) returns the required maximal speed squared
    // if the frog lands p units after the first wall.
    auto cost = [&](long double p) {
        // First jump:
        // from starting point to first wall is ds,
        // from first wall to landing point is p.
        long double first = min_v2_jump(ds, p, b1, t1);

        // Second jump:
        // from landing point to second wall is l - p,
        // from second wall to final point is df.
        long double second = min_v2_jump(l - p, df, b2, t2);

        // The frog needs one maximum speed that works for both jumps.
        return max(first, second);
    };

    // Ternary search interval for landing point p.
    // p must be between the walls, so 0 < p < l.
    long double lo = 0.0L, hi = l;

    // Perform many iterations for high precision.
    for(int iter = 0; iter < 300; iter++) {
        // First trisection point.
        long double m1 = lo + (hi - lo) / 3.0L;

        // Second trisection point.
        long double m2 = hi - (hi - lo) / 3.0L;

        // Since cost is unimodal, discard the worse third.
        if(cost(m1) < cost(m2)) {
            // Minimum lies to the left of m2.
            hi = m2;
        } else {
            // Minimum lies to the right of m1.
            lo = m1;
        }
    }

    // The best p is approximately the middle of the final interval.
    long double p = (lo + hi) / 2.0L;

    // cost(p) is speed squared, so output its square root.
    cout << sqrtl(cost(p)) << '\n';
}

// Program entry point.
int main() {
    // Speed up C++ I/O.
    ios_base::sync_with_stdio(false);

    // Untie cin from cout for faster input.
    cin.tie(nullptr);

    // Print fixed-point numbers with six digits after the decimal point.
    // This is enough for 1e-4 accuracy.
    cout << fixed << setprecision(6);

    // Process test cases until EOF.
    while(read()) {
        solve();
    }

    // Successful termination.
    return 0;
}
```

---

## 4. Python solution with detailed comments

```python
import sys
import math


def min_speed_squared_for_jump(d1, d2, b, t, g):
    """
    Compute the minimum required speed squared for one jump.

    The frog:
    - starts at height 0,
    - crosses the wall after horizontal distance d1,
    - lands at height 0 after another horizontal distance d2,
    - must cross the wall at some height h in [b, t].

    The derived formula is:

        v^2(h) = A / h + B * h

    where:

        A = g * d1 * d2 / 2
        B = g * (d1 + d2)^2 / (2 * d1 * d2)

    The unconstrained minimizing height is:

        h = d1 * d2 / (d1 + d2)

    Then we clamp h to the hole interval [b, t].
    """

    # Coefficient A.
    A = g * d1 * d2 / 2.0

    # Coefficient B.
    B = g * (d1 + d2) * (d1 + d2) / (2.0 * d1 * d2)

    # Best crossing height without hole restrictions.
    h0 = d1 * d2 / (d1 + d2)

    # Clamp crossing height into the allowed hole interval.
    h = max(b, min(t, h0))

    # Return the minimum possible speed squared.
    return A / h + B * h


def solve_case(b1, t1, b2, t2, l, ds, df, g):
    """
    Solve one test case.
    """

    def cost(p):
        """
        Required maximum speed squared if the frog lands p units
        after the first wall.
        """

        # First jump:
        # start -> first wall distance is ds,
        # first wall -> landing point distance is p.
        first = min_speed_squared_for_jump(ds, p, b1, t1, g)

        # Second jump:
        # landing point -> second wall distance is l - p,
        # second wall -> final point distance is df.
        second = min_speed_squared_for_jump(l - p, df, b2, t2, g)

        # The frog's maximal speed must handle both jumps.
        return max(first, second)

    # The landing point must be between the walls.
    lo = 0.0
    hi = l

    # Ternary search on the unimodal function cost(p).
    # 200 iterations are already more than enough for double precision.
    for _ in range(200):
        # First trisection point.
        m1 = lo + (hi - lo) / 3.0

        # Second trisection point.
        m2 = hi - (hi - lo) / 3.0

        # Discard the third that cannot contain the minimum.
        if cost(m1) < cost(m2):
            hi = m2
        else:
            lo = m1

    # Approximate optimal landing point.
    p = (lo + hi) / 2.0

    # cost(p) is speed squared, so return square root.
    return math.sqrt(cost(p))


def main():
    # Read all real numbers from input.
    data = list(map(float, sys.stdin.read().split()))

    # Each test case consists of exactly eight numbers.
    out = []

    for i in range(0, len(data), 8):
        # Extract one test case.
        b1, t1, b2, t2, l, ds, df, g = data[i:i + 8]

        # Solve it.
        ans = solve_case(b1, t1, b2, t2, l, ds, df, g)

        # Six digits after decimal are enough for 1e-4 accuracy.
        out.append(f"{ans:.6f}")

    # Print all answers.
    print("\n".join(out))


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

---

## 5. Compressed editorial

For a fixed landing point `p` between the two walls, the two jumps are independent.

For one jump, let:

- `d1` = distance from start to wall,
- `d2` = distance from wall to landing point,
- `h` = height where the frog crosses the wall.

The trajectory starts and ends at ground height, so:

```text
y(x) = a x (d1 + d2 - x)
```

At the wall:

```text
h = a d1 d2
```

Projectile motion gives:

```text
v^2 = g d1 d2 / (2h)
    + g h (d1 + d2)^2 / (2 d1 d2)
```

So:

```text
v^2(h) = A / h + B h
```

This convex function is minimized at:

```text
h = d1 d2 / (d1 + d2)
```

Clamp that height into the hole interval `[b, t]`, then evaluate `v^2`.

For landing point `p`:

```text
cost(p) = max(
    jump(ds, p, b1, t1),
    jump(l - p, df, b2, t2)
)
```

where `jump(...)` returns minimum speed squared.

The function `cost(p)` is unimodal on `(0, l)`, so use ternary search. The answer is:

```text
sqrt(min cost(p))
```

Use enough iterations, e.g. `200` or `300`, and print with at least four digits after the decimal point.