You are an ice cream man. You serve portions of ice cream filling a cone up to some level, and top it with two equally-sized flavored gelato balls. You are also generous, and want to put the largest balls you can. Both balls have to touch the main portion of the ice cream. The balls can touch, but not intersect. The balls can also touch, but not intersect the cone. What is the maximum possible radius of the ice cream balls?
Also, you live in 2D, so the cone is actually an isosceles triangle, and the balls are not spheres, but circles.
Given
The first line of input contains an integer $t$, the number of test cases ($1 <= t <= 10^4$). Each of the next $t$ lines contains three integers $h$, $ell$, $w$ ($1 <= h, ell, w <= 10^9$, $h > ell$), the description of a single test case.
For each test case, output a single real number, the maximum possible radius of the two balls. The output must be within $10^(-6)$ absolute or relative error of the correct answer.
## Input
The first line of input contains an integer $t$, the number of test cases ($1 <= t <= 10^4$). Each of the next $t$ lines contains three integers $h$, $ell$, $w$ ($1 <= h, ell, w <= 10^9$, $h > ell$), the description of a single test case.
## Output
For each test case, output a single real number, the maximum possible radius of the two balls. The output must be within $10^(-6)$ absolute or relative error of the correct answer.
[samples]
**Definitions**
Let $ C \in \mathbb{Z} $ be the number of scenarios.
For each scenario $ i \in \{1, \dots, C\} $, let $ (a_i, b_i) \in \mathbb{Z}^2 $ be the start point and $ (c_i, d_i) \in \mathbb{Z}^2 $ be the end point.
**Constraints**
1. $ 1 \leq C \leq 1000 $
2. For each $ i $, $ 0 \leq a_i, b_i, c_i, d_i \leq 10^{18} $
**Objective**
For each scenario $ i $, compute the minimum number of moves required to travel from $ (a_i, b_i) $ to $ (c_i, d_i) $, where:
- Each move is a unit step in one of the four cardinal directions (up, down, left, right).
- No two consecutive moves may be in the same direction.
Let $ \Delta x = |c_i - a_i| $, $ \Delta y = |d_i - b_i| $, and $ D = \Delta x + \Delta y $.
Then the minimum number of moves is:
$$
\begin{cases}
0 & \text{if } \Delta x = 0 \text{ and } \Delta y = 0 \\
\Delta x + \Delta y & \text{if } \Delta x = 0 \text{ or } \Delta y = 0 \text{ and } \Delta x + \Delta y \geq 1 \\
2 \cdot \min(\Delta x, \Delta y) + \max(\Delta x, \Delta y) & \text{if } \Delta x > 0 \text{ and } \Delta y > 0 \text{ and } |\Delta x - \Delta y| \leq 1 \\
2 \cdot \min(\Delta x, \Delta y) + \max(\Delta x, \Delta y) + 1 & \text{if } \Delta x > 0 \text{ and } \Delta y > 0 \text{ and } |\Delta x - \Delta y| \geq 2
\end{cases}
$$
But since consecutive moves cannot be in the same direction, the minimal path must alternate directions. The minimal number of moves is:
$$
\boxed{
\begin{cases}
0 & \text{if } \Delta x = 0 \text{ and } \Delta y = 0 \\
\Delta x + \Delta y & \text{if } \min(\Delta x, \Delta y) = 0 \\
2 \cdot \min(\Delta x, \Delta y) + \max(\Delta x, \Delta y) + \left( \max(\Delta x, \Delta y) - \min(\Delta x, \Delta y) > 1 \right) & \text{otherwise}
\end{cases}
}
$$
Equivalently, simpler and correct closed form:
Let $ dx = |c - a| $, $ dy = |d - b| $, $ m = \min(dx, dy) $, $ M = \max(dx, dy) $.
Then the answer is:
$$
\boxed{
\begin{cases}
dx + dy & \text{if } dx = 0 \text{ or } dy = 0 \\
2m + M & \text{if } M \leq m + 1 \\
2m + M + 1 & \text{if } M > m + 1
\end{cases}
}
$$