Alice and Bob are playing a fun game of tree tag.

The game is played on a tree of n vertices numbered from 1 to n. Recall that a tree on n vertices is an undirected, connected graph with n-1 edges.

Initially, Alice is located at vertex a, and Bob at vertex b. They take turns alternately, and Alice makes the first move. In a move, Alice can jump to a vertex with distance at most da from the current vertex. And in a move, Bob can jump to a vertex with distance at most db from the current vertex. The distance between two vertices is defined as the number of edges on the unique simple path between them. In particular, either player is allowed to stay at the same vertex in a move. Note that when performing a move, a player only occupies the starting and ending vertices of their move, not the vertices between them.

If after at most 10^{100} moves, Alice and Bob occupy the same vertex, then Alice is declared the winner. Otherwise, Bob wins.

Determine the winner if both players play optimally.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.

The first line of each test case contains five integers n,a,b,da,db (2≤ n≤ 10^5, 1≤ a,b≤ n, a≠ b, 1≤ da,db≤ n-1) — the number of vertices, Alice's vertex, Bob's vertex, Alice's maximum jumping distance, and Bob's maximum jumping distance, respectively.

The following n-1 lines describe the edges of the tree. The i-th of these lines contains two integers u, v (1≤ u, v≤ n, u≠ v), denoting an edge between vertices u and v. It is guaranteed that these edges form a tree structure.

It is guaranteed that the sum of n across all test cases does not exceed 10^5.

Output

For each test case, output a single line containing the winner of the game: "Alice" or "Bob".

Example

Input


4
4 3 2 1 2
1 2
1 3
1 4
6 6 1 2 5
1 2
6 5
2 3
3 4
4 5
9 3 9 2 5
1 2
1 6
1 9
1 3
9 5
7 9
4 8
4 3
11 8 11 3 3
1 2
11 9
4 9
6 5
2 10
3 2
5 9
8 3
7 4
7 10


Output


Alice
Bob
Alice
Alice

Note

In the first test case, Alice can win by moving to vertex 1. Then wherever Bob moves next, Alice will be able to move to the same vertex on the next move.

<image>

In the second test case, Bob has the following strategy to win. Wherever Alice moves, Bob will always move to whichever of the two vertices 1 or 6 is farthest from Alice.

<image>

import java.io.ByteArrayInputStream;
import java.io.File;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Scanner;


public class D_5 {
	public static void main(String[] args) throws Exception {
		new D_5().go();
	}

	int USE_FILE = 6;
	String INPUT = "4\r\n" +
			"4 3 2 1 2\r\n" +
			"1 2\r\n" +
			"1 3\r\n" +
			"1 4\r\n" +
			"6 6 1 2 5\r\n" +
			"1 2\r\n" +
			"6 5\r\n" +
			"2 3\r\n" +
			"3 4\r\n" +
			"4 5\r\n" +
			"9 3 9 2 5\r\n" +
			"1 2\r\n" +
			"1 6\r\n" +
			"1 9\r\n" +
			"1 3\r\n" +
			"9 5\r\n" +
			"7 9\r\n" +
			"4 8\r\n" +
			"4 3\r\n" +
			"11 8 11 3 3\r\n" +
			"1 2\r\n" +
			"11 9\r\n" +
			"4 9\r\n" +
			"6 5\r\n" +
			"2 10\r\n" +
			"3 2\r\n" +
			"5 9\r\n" +
			"8 3\r\n" +
			"7 4\r\n" +
			"7 10";


	void solve(int caseNr) {
		int n = in.nextInt();
		int a = in.nextInt();
		int b = in.nextInt();
		int da = in.nextInt();
		int db = in.nextInt();

		nodes = new Node[n+1];

		for (int i = 1; i < n+1; i++) {
			nodes[i] = new Node(i);
		}

		for (int i = 1; i <= n-1; i++) {
			int n1 = in.nextInt();
			int n2 = in.nextInt();
			// System.out.printf("(%d %d) \n", n1, n2);
			nodes[n1].c.add(nodes[n2]);
			nodes[n2].c.add(nodes[n1]);
		}



		// This method will find the tree-diameter starting from any node
		findDepth(1);
		findDepth(deepest_node.id);

		int tree_diameter = max_distance;

		findDistance(a,b);
		// System.out.println(ab_distance);



		//		for (int i = 1; i <= n; i++) {
		//			findDepth(i);
		//			findDepth(deepest_node.id);
		//			System.out.println(max_distance);
		//		}
		//		for (int i = 1; i <= n; i++) {
		//			findDistance(7,i);
		//			System.out.printf("distance from 8 to %2d = %4d\n", i, ab_distance);
		//		}

		//		out.printf("n=%d a=%d b=%d da=%d db=%d \n", n, a, b, da, db);
		//		for (int i = 1; i <= n; i++) {
		//			nodes[i].sortChildren();
		//			out.println(nodes[i]);
		//		}

		//		if (caseNr == 2 && print_wrong) {
		//			out.println("Bob");
		//			print_wrong = false;
		//			return;
		//		}


		if (ab_distance <= da) {
			out.println("Alice");
			return;
		}
		if (2*da >= tree_diameter) {
			out.println("Alice");
			return;
		}
		if (db <= 2*da) {
			out.println("Alice");
			return;
		}
		out.println("Bob");



		return;
	} // end of solve2




	// set for each testcase and otherwise unchanged
	Node[] nodes = null;

	// visited must be reset for each dfs and bfs traversals
	boolean[] visited = null;

	// these are set when doing getDepth(..)
	Node deepest_node;
	int max_distance;

	// these are set when searching for the distance to a target node
	int ab_distance;
	int target_node;
	boolean search_done;


	void findDistance(int n1, int n2) {
		int n = nodes.length;
		visited = new boolean[n+1];
		ab_distance = 0;
		target_node = n2;
		search_done = false;
		distance(nodes[n1], 0);
	}


	void distance(Node n, int depth) {
		if (search_done) return;
		visited[n.id] = true;

		if (n.id == target_node) {
			ab_distance = depth;
			search_done = true;
		}

		for (Node nod :n.c) {
			if (visited[nod.id]) continue;
			distance(nod, depth+1);
		}
	}

	// gets depth and assigns deepest_node
	void findDepth(int node_id) {
		int n = nodes.length;
		visited = new boolean[n+1];
		deepest_node = nodes[node_id];
		max_distance = 0;
		depth(nodes[node_id], 0);
	}


	void depth(Node n, int depth) {
		visited[n.id] = true;

		// System.out.println(depth);
		// System.out.println("n.id = "+n.id);

		if (depth > max_distance) {
			max_distance = depth;
			deepest_node = n;
		}

		for (Node nod :n.c) {
			if (visited[nod.id]) continue;
			depth(nod, depth+1);
		}
	}


	class Node implements Comparable<Node> {
		int id;
		List<Node> c = new ArrayList<>();

		public Node(int id) {
			this.id = id;
		}

		@Override
		public String toString() {
			String str_rtn = "";
			for (Node n : c) {
				str_rtn += String.format("%d > %d    ", id, n.id);
			}
			return str_rtn;
		}

		public void sortChildren() {
			Collections.sort(c);
		}

		@Override
		public int compareTo(Node n2) {
			return this.id - n2.id;
		}
	}



	// ---------------------------------------------------------------------------------------------- //

	boolean HAS_TESTCASES = true;
	void testCases() {
		int t = 1;
		if (HAS_TESTCASES) {
			t = in.nextInt();
		}
		for (int i=1; i<=t; i++) {
			solve(i);
		}
		out.flush();
	}



	// ---------------------------------------------------------------------------------------------- //


	void go() throws Exception {


		if (LOCAL && USE_FILE == 0) {
			//  needs to be set locally or expects input from terminal
			System.setIn(new ByteArrayInputStream(INPUT.getBytes()));
			in = new Scanner(System.in);
		}
		if (LOCAL && USE_FILE > 0) {
			String filename = String.format("in/cf%d.txt", USE_FILE);
			File inputFile = new File(filename);
			in = new Scanner(inputFile);
			out.printf("[%s]\n", filename);
			out.flush();
		}
		if (!LOCAL) {
			in = new Scanner(System.in);
		}

		long startTime = System.currentTimeMillis();
		testCases();
		if (LOCAL) {
			System.out.printf("[%dms]\n", System.currentTimeMillis()-startTime);
		}
		in.close();
	}

	static boolean LOCAL = System.getProperty("ONLINE_JUDGE") == null;
	Scanner in;
	PrintWriter out = new PrintWriter(System.out);

}




