F. Ode to the Bridge Buildertime limit per test3 secondsmemory limit per test512 megabytesinputstandard inputoutputstandard output You are given a 2D plane. Initially, two points $$$P_1$$$ and $$$P_2$$$ are located at $$$(0,0)$$$ and $$$(1,0)$$$, respectively, with a line segment connecting them. You can draw triangles using the following operation: Choose two points $$$A$$$ and $$$B$$$ that are directly connected by a line segment (i.e., $$$A$$$ and $$$B$$$ must be the endpoints of an existing segment). Choose two real numbers $$$x$$$ and $$$y$$$ such that $$$-2\cdot 10^4\le x, y\le 2\cdot 10^4$$$. Draw a new point $$$C$$$ at coordinates $$$(x,y)$$$. Then, draw line segments $$$AC$$$ and $$$BC$$$, forming triangle $$$\triangle ABC$$$. The length $$$l$$$ of every side of the new triangle $$$\triangle ABC$$$ must satisfy $$$0.5\le l\le 1$$$. Note that each operation draws one new point and two new line segments.Your task is to draw a point at coordinates $$$(p, q)$$$ using at most $$$m = \left\lceil 2\sqrt{p^2 + q^2}\right\rceil$$$ operations, where $$$\lceil x\rceil$$$ is the smallest integer greater than or equal to $$$x$$$. It is guaranteed that $$$p$$$ and $$$q$$$ are positive integers.Due to potential rounding errors: The constructed vertex only needs to be within $$$10^{-4}$$$ of $$$(p,q)$$$. For triangle side lengths, an absolute error of $$$10^{-8}$$$ is allowed (i.e., $$$0.5 - 10^{-8} \le l \le 1 + 10^{-8}$$$). Note that the target point may be created before the last operation (see the second test case).InputEach test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 1000$$$). The description of the test cases follows. The first line of each test case contains three integers $$$p$$$, $$$q$$$, and $$$m$$$ ($$$1\le p, q\le 10^4$$$, $$$m = \left\lceil 2\sqrt{p^2 + q^2}\right\rceil$$$) — the $$$x$$$ and $$$y$$$ coordinates of the target, and the maximum number of operations you are allowed to perform, respectively.It is guaranteed that the sum of $$$m$$$ over all test cases does not exceed $$$10^5$$$. OutputFor each test case, output a single integer $$$n$$$ ($$$0\le n\le m$$$) representing the number of operations used.Then, output $$$n$$$ lines describing the operations. The $$$i$$$-th of the next $$$n$$$ lines should contain four values: two integers $$$u$$$ and $$$v$$$ ($$$1\le u,v\le i+1,u\neq v$$$), the identifiers of the two chosen points, followed by two real numbers $$$x$$$ and $$$y$$$ ($$$-2\cdot 10^4\le x,y\le 2\cdot 10^4$$$), the coordinates of the new point.The identifiers are assigned as follows: The identifiers of point $$$P_1(0,0)$$$ and $$$P_2(1,0)$$$ are $$$1$$$ and $$$2$$$, respectively. The identifier of the point drawn during the $$$j$$$-th operation is $$$j+2$$$. If there are multiple valid solutions that use at most $$$m$$$ operations, you may output any of them.ExampleInput21 1 33 1 7Output2
1 2 0.5 0.8660254037844386
2 3 1 1
7
1 2 0.5 0.5
2 3 1.1339745962156 0.5
4 2 1.8660254037844 0.5
4 5 2 1
5 6 2.5 0.5
6 7 3 1
6 7 2.5 1
NoteIn the first case, you can perform at most $$$\lceil 2\sqrt{2}\rceil=3$$$ operations. The following solution uses only two operations: Choose points $$$A = P_1$$$ and $$$B = P_2$$$. Draw a new point $$$P_3$$$ at coordinates $$$\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$$$. The lengths of all sides of triangle $$$\triangle P_1P_2P_3$$$ are all $$$1$$$. Choose points $$$A = P_2$$$ and $$$B = P_3$$$. Draw a new point $$$P_4$$$ at coordinates $$$(1, 1)$$$. The lengths of the sides of triangle $$$\triangle P_2P_3P_4$$$ are $$$|P_2P_3|=|P_2P_4|=1$$$, $$$|P_3P_4|=2\sin(15^\circ)\approx0.5176\ge 0.5$$$. In the second case, you can perform at most $$$\left\lceil 2\sqrt{10}\right\rceil=7$$$ operations.In the solution shown below, the points $$$P_4$$$ and $$$P_5$$$ are at coordinates $$$\left(2-\frac{\sqrt{3}}{2},\frac{1}{2}\right)$$$ and $$$\left(1+\frac{\sqrt{3}}{2},\frac{1}{2}\right)$$$ respectively, and $$$|P_4P_6|=|P_2P_5|=1$$$. It can be verified that the length of every line segment lies within $$$[0.5,1]$$$.Note that the last operation is unnecessary, and the solution is still correct without the last operation.