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.

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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.*;
import java.util.*;
import static java.util.Collections.*;
import static java.lang.Math.*;
import java.util.stream.*;

import javax.sound.sampled.Mixer;

import static java.util.Arrays.*;

@SuppressWarnings("unchecked")
public class D_Tree_Tag {
    public static PrintWriter out;
    public static InputReader in;
    public static long MOD = (long)1e9+7;
    public static int n,a,b,da,db;
    public static Graph g;
    public static int dab;
    public static int mxdist;
    static class Graph
    {
        public int n;
        public ArrayList adjacency[];
        public Graph(int nodes)
        {
            n = nodes;
            adjacency = new ArrayList[nodes];
            for(int i=0;i<nodes;i++)
            {
                adjacency[i] = new ArrayList<Integer>();
            }
        }
        @Override
        public String toString()
        {
            String ret = "";
            for (ArrayList temp : adjacency) {
                ret+=(temp.toString()+"\n");
            }
            return ret;
        }
        public void addEdge(int from, int to)
        {
            adjacency[from].add(to);
            adjacency[to].add(from);
        }
    }
    static class Pair{
        public int a,b;
        public Pair(int a, int b){
            this.a=a; this.b=b;
        }
    }
    public static void dist(){
        int[] vis = new int[n]; fill(vis,0);
        LinkedList<Pair> l = new LinkedList<Pair>();
        l.add(new Pair(a,0));
        while(!l.isEmpty()){
            Pair cur = l.pollFirst();
            int v = cur.a, ct = cur.b;
            vis[v] = 1;
            if(v==b) {dab = ct; return;}
            for(int neigh : (ArrayList<Integer>)g.adjacency[v]){
                if(vis[neigh]==0) l.add(new Pair(neigh,ct+1));
            }
        }
    }
    // https://www.geeksforgeeks.org/diameter-tree-using-dfs/#:~:text=We%20start%20DFS%20from%20a,the%20diameter%20of%20the%20tree.
    // Used to track farthest node. 
    static int x; 
    static int maxCount; 
    static List<Integer> adj[]; 
      
    // Sets maxCount as maximum distance  
    // from node 
    static void dfsUtil(int node, int count,  
                         boolean visited[], 
                       List<Integer> adj[]) 
    { 
        visited[node] = true; 
        count++; 
          
        List<Integer> l = adj[node]; 
        for(Integer i: l) 
        { 
            if(!visited[i]){ 
                if (count >= maxCount) { 
                    maxCount = count; 
                    x = i; 
                } 
                dfsUtil(i, count, visited, adj); 
            } 
        } 
    } 
       
    // The function to do DFS traversal. It uses  
    // recursive dfsUtil() 
    static void dfs(int node, int n, List<Integer> 
                                       adj[]) 
    { 
        boolean[] visited = new boolean[n + 1]; 
        int count = 0; 
       
        // Mark all the vertices as not visited 
        Arrays.fill(visited, false); 
       
        // Increment count by 1 for visited node 
        dfsUtil(node, count + 1, visited, adj); 
          
    } 
       
    // Returns diameter of binary tree represented 
    // as adjacency list. 
    static int diameter(List<Integer> adj[], int n) 
    { 
        maxCount = Integer.MIN_VALUE; 
       
        /* DFS from a random node and then see 
        farthest node X from it*/
        dfs(1, n, adj); 
       
        /* DFS from X and check the farthest node 
        from it */
        dfs(x, n, adj); 
       
        return maxCount; 
    } 
    public static int dfs2(int v, int[] vis){
        ArrayList<Integer> l = new ArrayList<Integer>();
        vis[v] = 1;
        int count = 0;
        for(int neigh : (ArrayList<Integer>)g.adjacency[v]){
            if(vis[neigh]==0) {count++; l.add(dfs2(neigh,vis));}
        }
        if(count==0) return 0;
        if(count==1) {
            // out.println("from i="+v+",mxdist="+l.get(0));
            mxdist = max(mxdist, 1+l.get(0)); return (1+l.get(0));}
        sort(l);
        // out.println("i="+v+",l="+l+",val = "+(2 + l.get(l.size()-1) + l.get(l.size()-2)));
        
        // out.println("before updating, mxdist:"+mxdist);
        mxdist = max(mxdist,2 + l.get(l.size()-1) + l.get(l.size()-2));
        // out.println("updated mxdist:"+mxdist);
        // out.println("for v="+v+",val:"+(1+l.get(l.size()-1)));
        return 1+l.get(l.size()-1);
    }
    public static void main(String[] args)throws IOException {
        in = new InputReader(System.in);
        out = new PrintWriter(System.out);
        
        int cases = in.nextInt();
        
        for(int t = 0; t < cases; t++){
            n = in.nextInt();
            a = in.nextInt()-1;
            b = in.nextInt()-1;
            da = in.nextInt();
            db = in.nextInt();
            g = new Graph(n);
            for(int i=0;i<n-1;i++){
                int u = in.nextInt()-1;
                int v=  in.nextInt()-1;
                g.addEdge(u, v);
            }
            // dab=-1;
            dist();
            mxdist = -1;
            int[] vis = new int[n]; fill(vis,0);
            int bhaiskiaankh = dfs2(a,vis);
            mxdist = max(mxdist,bhaiskiaankh);
            // mxdist = max(mxdist,dfs2(a,vis));
            // out.println("initially: "+mxdist);
            int other = mxdist;
            mxdist = diameter(g.adjacency, n)-1;
            if(other!=mxdist){
                out.println("other:"+other);
                out.println("mxdist:"+mxdist);
                out.println("FUCK");
            }
            // mxdist = -1;
            // mxdist = -1;
            // mxdist = max(mxdist,dfs2(a,vis));
            // out.println("mxdist="+mxdist);
            if(dab<=da){
                out.println("Alice");
            }
            else{
                if(da>=db) out.println("Alice");
                else{
                    if(db>(2*da) && mxdist>(2*da)){
                        out.println("Bob");
                    }
                    else out.println("Alice");
                }
            }
            // out.println("dab:"+dab);
            // out.println("dist:"+mxdist);
        }

        out.close();
    }
    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public int[] nextArrI(int n) {
            int[] a=new int[n];
            for(int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }
        
        public long[] nextArrL(int n) {
            long[] a=new long[n];
            for (int i=0; i<n; i++) a[i]=nextLong();
            return a;
        }
    }

}