## 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. 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;
};

long double b1, t1, b2, t2, l, ds, df, g;

bool read() {
    cin >> b1 >> t1 >> b2 >> t2 >> l >> ds >> df >> g;
    return !cin.fail();
}

void solve() {
    // The free variables are the landing point p in (0, l) between the
    // walls and the two hole-crossing heights h_1 in [b1, t1] and h_2 in
    // [b2, t2]. The jumper has a single max-speed v that must cover both
    // jumps, so for a given p the smallest feasible v is max(v1(p), v2(p))
    // with v_i(p) the minimum speed for the i-th jump. For a fixed p the
    // two jumps are independent and we minimise each v_i over its h_i in
    // closed form; that leaves p, which couples the jumps (moving p right
    // lengthens jump 1 and shortens jump 2) and we handle by ternary
    // searching
    //
    //     min over p of max(v1(p), v2(p)).
    //
    // Consider a jump from (0, 0) over the hole at (d1, h) to (d1 + d2, 0).
    // We pick coordinates so launch and landing both have y = 0, so the
    // trajectory is a parabola with roots at x = 0 and x = R := d1 + d2.
    // Any such parabola has the form
    //
    //     y(x) = a * x * (R - x),
    //
    // which already bakes in the two ground crossings; only the single
    // shape parameter a is left to fix. We fix it from the hole point:
    // y(d1) = a * d1 * d2 = h, so a = h / (d1 * d2).
    //
    // To turn this geometric parabola into a physical trajectory, write
    // the projectile motion x(t) = v_x * t, y(t) = v_y * t - g * t^2 / 2,
    // eliminate t via t = x / v_x:
    //
    //     y(x) = (v_y / v_x) * x  -  (g / (2 * v_x^2)) * x^2.
    //
    // Matching the x^2 coefficient with a * x * (R - x) = a * R * x - a * x^2
    // gives a = g / (2 * v_x^2), i.e. v_x^2 = g / (2 * a). And matching the
    // linear coefficient gives v_y / v_x = a * R, i.e. tan(theta) = a * R.
    // Then
    //
    //     v^2 = v_x^2 + v_y^2 = v_x^2 * (1 + tan^2 theta)
    //         = g / (2 * a) + g * a * R^2 / 2
    //         = A / h + B * h,
    //
    //     A = g * d1 * d2 / 2,   B = g * (d1 + d2)^2 / (2 * d1 * d2).
    //
    // f(h) = A / h + B * h is convex on h > 0 with unconstrained minimum
    // at h_star = sqrt(A / B) = d1 * d2 / (d1 + d2) and minimum value
    // g * (d1 + d2). The hole forces h in [b, t], so the optimal h is
    // clamp(h_star, b, t) and v_i(p)^2 = f at that h.
    //
    // Each v_i(p) is U-shaped in p (constraint-bound for extreme p,
    // unconstrained = g * (d1 + d2) in the middle), and v1, v2 move in
    // opposite directions with p, so max(v1, v2) is unimodal and ternary
    // search on p in (0, l) finds the optimum. -1 is never needed: every
    // jump is feasible for sufficiently large v since b > 0.

    auto min_v2_jump = [](long double d1, long double d2, long double b,
                          long double t) {
        long double A = g * d1 * d2 / 2.0L;
        long double B = g * (d1 + d2) * (d1 + d2) / (2.0L * d1 * d2);
        long double h = max(b, min(t, d1 * d2 / (d1 + d2)));
        return A / h + B * h;
    };

    auto cost = [&](long double p) {
        return max(min_v2_jump(ds, p, b1, t1), min_v2_jump(l - p, df, b2, t2));
    };

    long double lo = 0.0L, hi = l;
    for(int iter = 0; iter < 300; iter++) {
        long double m1 = lo + (hi - lo) / 3.0L;
        long double m2 = hi - (hi - lo) / 3.0L;
        if(cost(m1) < cost(m2)) {
            hi = m2;
        } else {
            lo = m1;
        }
    }

    cout << sqrtl(cost((lo + hi) / 2.0L)) << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout << fixed << setprecision(6);

    while(read()) {
        solve();
    }

    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.