### Problem

Given a vertex-colored tree with N nodes, can it be drawn in a 2D plane with a line of symmetry?

Formally, a tree is line-symmetric if each vertex can be assigned a location in the 2D plane such that:

• All locations are distinct.
• If vertex vi has color C and coordinates (xi, yi), there must also be a vertex vi' of color C located at (-xi, yi) -- Note if xi is 0, vi and vi' are the same vertex.
• For each edge (vi, vj), there must also exist an edge (vi', vj').
• If edges were represented by straight lines between their end vertices, no two edges would share any points except where adjacent edges touch at their endpoints.

### Input

The first line of the input gives the number of test cases, T. T test cases follow.

Each test case starts with a line containing a single integer N, the number of vertices in the tree.

N lines then follow, each containing a single uppercase letter. The i-th line represents the color of the i-th node.

N-1 lines then follow, each line containing two integers i and j (1 ≤ i < jN). This denotes that the tree has an edge from the i-th vertex to the j-th vertex. The edges will describe a connected tree.

### Output

For each test case, output one line containing "Case #x: y", where x is the case number (starting from 1) and y is "SYMMETRIC" if the tree is line-symmetric by the definition above or "NOT SYMMETRIC" if it isn't.

1 ≤ T ≤ 100.

2 ≤ N ≤ 12.

2 ≤ N ≤ 10000.

### Sample

 Input Output ```3 4 R G B B 1 2 2 3 2 4 4 R G B Y 1 2 2 3 2 4 12 Y B Y G R G Y Y B B B R 1 3 1 9 1 10 2 3 3 7 3 8 3 11 4 8 5 7 6 7 8 12 ``` ```Case #1: SYMMETRIC Case #2: NOT SYMMETRIC Case #3: SYMMETRIC ```

The first case can be drawn as follows: No arrangement of the second case has a line of symmetry: One way of drawing the third case with a symmetry line is as follows: Points Correct Attempted
7pt 22 24
18pt 15 22