#ARC161E. [ARC161E] Not Dyed by Majority (Cubic Graph)

[ARC161E] Not Dyed by Majority (Cubic Graph)

Score : 800800 points

Problem Statement

You are given a connected simple undirected graph with NN vertices and 32N\displaystyle\frac{3}{2}N edges, where NN is a positive even number. The vertices are labeled from 11 to NN, and the ii-th edge connects vertex AiA_i and vertex BiB_i. Moreover, for every vertex, the number of incident edges is exactly 3.

You will color each vertex of the given graph either black ( B ) or white ( W ). Here, the string obtained by arranging the colors ( B or W ) of the vertices in the order of vertex labels is called a color sequence.

Determine whether there is a color sequence that cannot result from performing the following operation once when all vertices are colored, and if there is such a color sequence, find one.

Operation: For each vertex k=1,2,,Nk = 1, 2, \dots, N, let CkC_k be the color that occupies the majority (more than half) of the colors of the vertices connected to kk by an edge. For every vertex kk, change its color to CkC_k simultaneously.

There are TT test cases to solve.

Constraints

  • T1T \geq 1
  • N4N \geq 4
  • The sum of NN over the test cases in each input file is at most 5×1045 \times 10^4.
  • NN is an even number.
  • $1 \leq A_i < B_i \leq N \ \ \left(1 \leq i \leq \displaystyle\frac{3}{2}N\right)$
  • $(A_i, B_i) \neq (A_j, B_j) \ \ \left(1 \leq i < j \leq \displaystyle\frac{3}{2}N\right)$
  • The given graph is connected.
  • Each vertex k (1kN)k \ (1 \leq k \leq N) appears exactly 3 times as $A_i, B_i \ \left(1 \leq i \leq \displaystyle\frac{3}{2}N\right)$.

Input

The input is given from Standard Input in the following format:

TT

case1\mathrm{case}_1

case2\mathrm{case}_2

\vdots

caseT\mathrm{case}_T

Each test case casei (1iT)\mathrm{case}_i \ (1 \leq i \leq T) is in the following format:

NN

A1A_1 B1B_1

A2A_2 B2B_2

\vdots

A32NA_{\frac{3}{2}N} B32NB_{\frac{3}{2}N}

Output

Print TT lines. For the ii-th line, if there is a color sequence that cannot result from performing the operation for the ii-th test case, print such a color sequence; otherwise, print -1. If there are multiple color sequences that cannot result from performing the operation, any of them will be considered correct.

2
4
1 2
1 3
1 4
2 3
2 4
3 4
10
1 2
1 3
1 4
2 3
2 4
3 5
4 5
5 6
6 7
6 8
7 9
7 10
8 9
8 10
9 10
BWWW
BWWWBWWWBB

Let us consider the first test case. For the color of vertex 11 to be B, at least two of the colors of vertices 2,3,42, 3, 4 must be B before the operation. Then, for at least one of the vertices 2,3,42, 3, 4, at least two of the colors of the vertices connected to that vertex by an edge are B, so the color of that vertex after the operation will be B. Therefore, the color sequence BWWW cannot result from performing the operation.