Alphonse is assembling an alphabet tree and arranging some adventures along the way.

An alphabet tree is an unrooted tree with (N) nodes (numbered from (1) to (N)) and (N - 1) edges. Initially, the (i)th edge connects nodes (A_i) and (B_i) in both directions, and is labeled with a uppercase letter (C_i). Two edges incident to a common node are always labeled with different letters.

Alphonse has (Q) events to process, the (i)th of which is one of two types:

1 (U_i) (L_i): Add a new node to the tree by connecting it to node (U_i) with a new edge labeled with uppercase letter (L_i). Newly added nodes are numbered with integers starting from (N + 1) in the order they're added.
2 (U_i) (K_i) (S_i): Print the final node Alphonse will end up at if he:
- Starts a journey at node (U_i).
- Repeatedly traverses a previously untraversed edge (on this journey). If his current node has multiple untraversed edges, he picks the edge labeled with the letter that comes earliest in the string (S_i).
- Ends the journey once there are no more untraversed edges at the current node, or (K_i) edges have been traversed on the journey.

Please help Alphonse determine where each journey will take him.

Constraints

(1 \le T \le 20) (2 \le N \le 300{,}000) (1 \le Q \le 300{,}000) (1 \le A_i, B_i \le N) (A_i \ne B_i) (C_i, L_i, S_{i,j} \in {)'A', (\ldots), 'Z'(}) (1 \le U_i, K_i \le 600{,}000)

The sum of (N) over all test cases is at most (1{,}100{,}000). The sum of (Q) over all test cases is at most (1{,}100{,}000).

For each event, it is guaranteed that:

(U_i) is a valid node in the tree at the time of the event.
(L_i) is different from all existing labels of edges incident to (U_i) at the time of the event.
(S_i)'s letters are distinct, and are a superset of all edge labels in the tree at the time of the event.

Input Format

Input begins with a single integer (T), the number of test cases. For each test case, there is first a line containing a single integer (N). Then, (N - 1) lines follow, the (i)th of which contains space-separated integers (A_i) and (B_i), followed by a space, followed by (C_i). Then, there is a line containing the single integer (Q). Then, (Q) lines follow, the (i)th of which is either 1 (U_i) (L_i) or 2 (U_i) (K_i) (S_i).

Output Format

For the (i)th test case, print a single line containing "Case #i: ", followed by space-separated integers, the answers to all type-2 events.

Sample Explanation

The first sample case's alphabet tree is depicted above, and yields the following journeys:

Event (1) traverses (1 \stackrel{\texttt{M}}{\longrightarrow} 2 \stackrel{\texttt{E}}{\longrightarrow} 6) (ended after no more edges from node (6))
Event (2) traverses (3 \stackrel{\texttt{E}}{\longrightarrow} 1 \stackrel{\texttt{T}}{\longrightarrow} 4 \stackrel{\texttt{A}}{\longrightarrow} 9) (ended after (K_2 = 3) steps)
Event (3) traverses (9 \stackrel{\texttt{A}}{\longrightarrow} 4 \stackrel{\texttt{T}}{\longrightarrow} 1 \stackrel{\texttt{M}}{\longrightarrow} 2) (ended after (K_3 = 3) steps)
Event (4) traverses (8 \stackrel{\texttt{M}}{\longrightarrow} 3 \stackrel{\texttt{E}}{\longrightarrow} 1 \stackrel{\texttt{T}}{\longrightarrow} 4 \stackrel{\texttt{A}}{\longrightarrow} 9) (ended after no more edges from node (9))
Event (6) traverses (8 \stackrel{\texttt{T}}{\longrightarrow} 10) (ended after no more edges from node (10))

The second and third sample cases are depicted below.

In the second case, the journeys are:

Event (1) traverses (1 \stackrel{\texttt{P}}{\longrightarrow} 5 \stackrel{\texttt{C}}{\longrightarrow} 2) (ended after (K_1 = 2) steps)
Event (2) traverses (4 \stackrel{\texttt{P}}{\longrightarrow} 2 \stackrel{\texttt{C}}{\longrightarrow} 5 \stackrel{\texttt{P}}{\longrightarrow} 1) (ended after (K_2 = 3) steps)
Event (4) traverses (4 \stackrel{\texttt{P}}{\longrightarrow} 2 \stackrel{\texttt{C}}{\longrightarrow} 5 \stackrel{\texttt{U}}{\longrightarrow} 6) (ended after (K_4 = 3) steps)
Event (5) traverses (3 \stackrel{\texttt{U}}{\longrightarrow} 2 \stackrel{\texttt{P}}{\longrightarrow} 4) (ended after no more edges from node (4))

In the third case, the journeys are:

Event (1) traverses (3 \stackrel{\texttt{C}}{\longrightarrow} 2 \stackrel{\texttt{A}}{\longrightarrow} 1) (ended after (K_1 = 2) steps)
Event (2) traverses (4 \stackrel{\texttt{E}}{\longrightarrow} 2 \stackrel{\texttt{A}}{\longrightarrow} 1) (ended after (K_2 = 2) steps)
Event (3) traverses (1 \stackrel{\texttt{A}}{\longrightarrow} 2) (ended after (K_3 = 1) steps)