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.*; 
import java.util.*; 
//import javafx.util.*; 
import java.math.*;
//import java.lang.*;
public class Main 
{ 
  //   static long mod=1000000007;
     static HashSet<Integer> adj[];
   //  static boolean vis[];
  //   static int n[];
  //   static ArrayList<Integer> arr[];
  // //  static long ans=0;
  //   static boolean vis[][];
  //   static final long oo=(long)1e18;
    // static int par[];
    // static long child[];
    // static HashMap<String,Long> hm=new HashMap<>();
    static int ver=0;static int edge=0;
    static int node, diam;
    static int [] d;
     public static void main(String[] args) throws IOException {  
        PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        br = new BufferedReader(new InputStreamReader(System.in));
        int test=nextInt();
       // int test=1;
        outer: while(test--!=0){
            int n=nextInt();
            int a=nextInt()-1;
            int b=nextInt()-1;
            int da=nextInt();
            int db=nextInt();
            d=new int[n];
            adj=new HashSet[n];
            for(int i=0;i<n;i++){
                adj[i]=new HashSet<>();
            }
            for(int i=0;i<n-1;i++){
                int aa=nextInt()-1;
                int bb=nextInt()-1;
                adj[aa].add(bb);
                adj[bb].add(aa);
            }
            if(db<=da*2||a==b){
                pw.println("Alice");
                continue outer;
            }
            Deque<Integer> dq=new ArrayDeque<>();
            boolean vis[]=new boolean[n];
            dq.add(a);
            vis[a]=true;
            int dist=0;
            out:while(!dq.isEmpty()){
                int len=dq.size();
                for(int i=0;i<len;i++){
                    int v=dq.remove();
                    if(v==b){
                        break out;
                    }
                    for(int ver:adj[v]){
                        if(!vis[ver]){
                            dq.add(ver);
                            vis[ver]=true;
                        }
                    }
                }
                dist++;
            } 
            if(dist<=da){
                pw.println("Alice");
                continue outer;
            }
            findDiam(0);
            int rad=(diam+1)/2;
           // pw.println(rad);
            if(rad<=da){
                pw.println("Alice");
            }
            else{
                pw.println("Bob");    
            }


        }
           
        pw.close();
    }
    static void findDiam(int u){
        node = -1;
        diam = 0;
        dfs(u, -1, 0);
        diam = 0;
        dfs(node, -1, 0);
        d[u] = diam;
    }
    static void dfs(int u, int p, int l){
        if(l >= diam){
            diam = l;
            node = u;
        }
        for(int v:adj[u])
            if(v != p)
                dfs(v, u, l+1);
    }
    
    
    
    static class compare implements Comparable<compare>{
         String s;
         compare(String s){
            this.s=s;
        }
        public  int compareTo(compare t){
            return compareVersion(s,t.s);
        }
        public  int compareVersion(String version1, String version2) {
            String[] nums1 = version1.split("\\.");
            String[] nums2 = version2.split("\\.");
            int n1 = nums1.length, n2 = nums2.length;

            // compare versions
            int i1, i2;
            for (int i = 0; i < Math.max(n1, n2); ++i) {
              i1 = i < n1 ? Integer.parseInt(nums1[i]) : 0;
              i2 = i < n2 ? Integer.parseInt(nums2[i]) : 0;
              if (i1 != i2) {
                return i1 > i2 ? 1 : -1;
              }
            }
            // the versions are equal
            return 0;
        }
    }
  
    
    static class Pair{
        int a;int b;
        Pair(int a,int b){
          //  this.index=index;
            this.a=a;
            this.b=b;
            
            //this.z=z;
        }   
    }
   static void update(long tree[],int idx,long val){
        //println(idx);
        while(idx<1000006){
            
            tree[idx]+=val;
            idx+=(idx)&(-idx);
        }
    }   
    static long sum(long tree[],int idx){
        long sum=0;
        while(idx>0){
           
            sum+=tree[idx];
            idx-=(idx)&(-idx);
        }
        return sum;
    }
    public static BufferedReader br;
    public static StringTokenizer st;
    public static String next() {
        while (st == null || !st.hasMoreTokens()) {
       try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                throw new RuntimeException(e);
            }
        }
        return st.nextToken();
    }
 
    public static Integer nextInt() {
        return Integer.parseInt(next());
    }
    public static Long nextLong() {
        return Long.parseLong(next());
    }
 
    public static Double nextDouble() {
        return Double.parseDouble(next());
    }
    // static class Pair{
    //     int x;int y;
    //     Pair(int x,int y,int z){
    //         this.x=x;
    //         this.y=y;
    //        // this.z=z;
    //      //   this.z=z;
    //       //  this.i=i;
    //     }
    // }
    // static class sorting implements Comparator<Pair>{
    //     public int compare(Pair a,Pair b){
    //         //return (a.y)-(b.y);
    //         if(a.y==b.y){
    //             return -1*(a.z-b.z);
    //         }
    //         return (a.y-b.y);
    //     }
   // }

    static long ncr(long n,long r){
        
        if(r==0)
            return 1l;
        long val=ncr(n-1,r-1);
        val=(n*val);
        val=(val/r);
        return val;
    }

    public static int[] na(int n)throws IOException{
        int[] a = new int[n];
        for(int i = 0;i < n;i++)a[i] = nextInt();
        return a;
    }
    static class query implements Comparable<query>{
        int l,r,idx,block;
        static int len;
        query(int l,int r,int i){
            this.l=l;
            this.r=r;
            this.idx=i;
            this.block=l/len;
        }   
        public int compareTo(query a){
            return block==a.block?r-a.r:block-a.block;
        }
    }
    
    // static class sorting implements Comparator<Pair>{  
    //     public int compare(Pair a1,Pair a2){  
    //         if(o1.a==o2.a)
    //             return (o1.b>o2.b)?1:-1;  
    //         else if(o1.a>o2.a)  
    //             return 1;  
    //         else  
    //             return -1;  
    //     }
    // }  
    static boolean isPrime(int n) { 
        if (n <= 1) 
            return false; 
        if (n <= 3) 
            return true; 
        if (n % 2 == 0 ||  
            n % 3 == 0) 
            return false; 
      
        for (int i = 5; 
                 i * i <= n; i = i + 6) 
            if (n % i == 0 || 
                n % (i + 2) == 0) 
                return false; 
      
        return true; 
    } 
    static int gcd(int a, int b) { 
      if (b == 0) 
        return a; 
      return gcd(b, a % b);  
    }  
    // To compute x^y under modulo m 
    static long power(long x, long y, long m){ 
        if (y == 0) 
            return 1;      
        long p = power(x, y / 2, m) % m; 
        p = (p * p) % m; 
      
        if (y % 2 == 0) 
            return p; 
        else
            return (x * p) % m; 
    }
    static long fast_pow(long base,long n,long M){
        if(n==0)
           return 1;
        if(n==1)
        return base;
        long halfn=fast_pow(base,n/2,M);
        if(n%2==0)
            return ( halfn * halfn ) % M;
        else
            return ( ( ( halfn * halfn ) % M ) * base ) % M;
    }
       static long modInverse(long n,long M){
        return fast_pow(n,M-2,M);
    }
    // (1,1)   

    //     (3,2)   (3,5)

} 