**Problem Link:** contest, practice

**Difficulty:** Cakewalk

**Pre-requisites:** Geometry, Implementation

**Problem:**

We are given two numbers **A** and **B**. Our task is to determine the minimal and the maximal possible value of number **C** thus exists a non-obtuse triangle with the lengths of the sides equal to **A**, **B** and **C**.

It’s also guaranteed, that **A < B**.

**Explanation:**

It was the easiest problem of the contest.

Since **A < B**, the only angles, that could be obtuse, are the angles between sides **A** and **B** or **A** and **C**.

So, the minimal possible value of **C** is reached when the angle between sides **A** and **C** is right(equals to 90 degrees).

Also, the maximal possible value of **C** is reached when the angle between sides **A** and **B** is right(equals to 90 degrees).

The first value **C _{min} = sqrt( B^{2} - A^{2} )**;

The second value **C _{max} = sqrt( B^{2} + A^{2} )**.

The total complexity is **O(1)** per testcase.

**Setter’s Solution:** link

**Tester’s Solution:** link