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.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Queue;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;

import javax.sound.sampled.ReverbType;

import java.util.*;
import java.io.*;

public class Main {
	static ArrayList<Integer> adj[];
	// static PrintWriter out = new PrintWriter(System.out);

	static int[][] notmemo;
	static int k;
	static int[] a;
	static int b[];
	static int m;

	static class Pair implements Comparable<Pair> {
		int d;
		int s;

		public Pair(int b, int l) {
			d = b;
			s = l;
		}

		@Override
		public int compareTo(Pair o) {
			if (o.d - o.s > this.d - this.s) {
				return 1;
			} else if (o.d - o.s < this.d - this.s) {
				return -1;
			} else
				return 0;

		}

	}

	static Pair s1[];
	static ArrayList<Pair> adjlist[];
	// static char c[];
	public static long mod = (long) (1e9 + 7);
	static int V;
	static long INF = (long) 1E16;
	static int n;
	static char c[];
	static int d[];
	static int z;
	static Pair p[];
static int R;
static int C;
static int K;
static long grid[][];
	public static void main(String args[]) throws Exception {
		Scanner sc = new Scanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int t=sc.nextInt();
		while(t-->0) {
		int n=sc.nextInt();
		int a=sc.nextInt()-1;
		int b=sc.nextInt()-1;
		int da=sc.nextInt();
		int db=sc.nextInt();
		adj=new ArrayList[n];
		for (int i = 0; i <n; i++) {
			adj[i]=new ArrayList<>();
		}
		int temp=n;
		while(temp-->1) {
			int u=sc.nextInt()-1;
			int v=sc.nextInt()-1;
			adj[u].add(v);
			adj[v].add(u);
		}
		max=0;
		node=0;
	vis=new boolean[n];
	dfs1(node,1);
	vis=new boolean[n];
	dfs1(node,1);
	vis=new boolean[n];
	int[] dist = new int[n];
	Arrays.fill(dist, (int) 1e9);
	PriorityQueue<Integer> pq = new PriorityQueue<Integer>();
	dist[a] = 0;
	int ans=0;
	pq.add(a);
	while(!pq.isEmpty()) {
		int cur=pq.remove();
		if(cur==b) {
			ans=dist[cur];
			break;
		}
		for(int nxt:adj[cur]) {
			if(dist[cur]+1<dist[nxt]) {
				pq.add(nxt);
				dist[nxt] =dist[cur]+1;
			}
		}
	}
	
	if(max>da*2&&db>(da*2)&&ans>da) {
		System.out.println("Bob");
	}
	else {
		System.out.println("Alice");
	}
		}
	}
	static int max=0;
	static int node;
	static int tar;
	
	static void dfs1(int u,int count) {
	vis[u]=true;
	for(int v:adj[u]) {
		if(!vis[v]) {
			if(count>max) {
				max=count;
				node=v;
			}
			dfs1(v,count+1);
		}
	}
	}
	
	
	
	static  class FenwickTree { // one-based DS

		int n;
		int[] ft;
		
		FenwickTree(int size) { n = size; ft = new int[n+1]; }
		
		int rsq(int b) //O(log n)
		{
			int sum = 0;
			while(b > 0) { sum += ft[b]; b -= b & -b;}		//min?
			return sum;
		}
		
		int rsq(int a, int b) { return rsq(b) - rsq(a-1); }	
		
		void point_update(int k, int val)	//O(log n), update = increment
		{
			while(k <= n) { ft[k] += val; k += k & -k; }		//min?
		}
	}
	static ArrayList<Integer> euler=new ArrayList<>();
	

	static ArrayList<Integer> arr;

	
static long total[];
static TreeMap<Integer,Integer> map1;

		static int zz;
//static int dp(int idx,int left,int state) {
//if(idx>=k-((zz-left)*2)||idx+1==n) {
//	return 0;
//}
//if(memo[idx][left][state]!=-1) {
//	return memo[idx][left][state];
//}
//int ans=a[idx+1]+dp(idx+1,left,0);
//if(left>0&&state==0&&idx>0) {
//	ans=Math.max(ans,a[idx-1]+dp(idx-1,left-1,1));
//}
//return memo[idx][left][state]=ans;
//}21
static HashMap<Integer,Integer> map;
static int maxa=0;
static int ff=123123;

static int[][][] memo;


	static long modmod=998244353;
	static int dx[]= {1,-1,0,0};
	 static int dy[]= {0,0,1,-1};	
	static class BBOOK implements Comparable<BBOOK>{
		int t;
		int alice;
		int bob;
		public BBOOK(int x,int y,int z) {
		t=x;
		alice=y;
		bob=z;
		}
		@Override
		public int compareTo(BBOOK o) {
			return this.t-o.t;
		}
	}
	private static long lcm(long a2, long b2) {
	return (a2*b2)/gcd(a2,b2);
	}
	static class Edge implements Comparable<Edge>
	{
		int node;long cost ; long time; 
		
		Edge(int a, long b,long c) { node = a;	cost = b; time=c; }
		
		public int compareTo(Edge e){ return Long.compare(time,e.time);	}
	}
	
        static void sieve(int N)	// O(N log log N) 
	{
		isComposite = new int[N+1];					
		isComposite[0] = isComposite[1] = 1;			// 0 indicates a prime number
		primes = new ArrayList<Integer>();
 
		for (int i = 2; i <= N; ++i) 					//can loop till i*i <= N if primes array is not needed O(N log log sqrt(N)) 
			if (isComposite[i] == 0) 					//can loop in 2 and odd integers for slightly better performance
			{
				primes.add(i);
				if(1l * i * i <= N)
					for (int j = i * i; j <= N; j += i)	// j = i * 2 will not affect performance too much, may alter in modified sieve
						isComposite[j] = 1;
			}   
	}
    static TreeSet<Integer> factors;
    static ArrayList<Integer> primeFactors(int N)		// O(sqrt(N) / ln sqrt(N))
	{
		ArrayList<Integer> factors = new ArrayList<Integer>();		//take abs(N) in case of -ve integers
		int idx = 0, p = primes.get(idx);

		while(1l*p * p <= N)
		{
			while(N % p == 0) { factors.add(p); N /= p; }
			p = primes.get(++idx);
		}

		if(N != 1)						// last prime factor may be > sqrt(N)
			factors.add(N);				// for integers whose largest prime factor has a power of 1
		return factors;
	}

	static class UnionFind {
		int[] p, rank, setSize;
		int numSets;
		int max[];

		public UnionFind(int N) {
			p = new int[numSets = N];
			rank = new int[N];
			setSize = new int[N];
			for (int i = 0; i < N; i++) {
				p[i] = i;
				setSize[i] = 1;
			}
		}

		public int findSet(int i) {
			return p[i] == i ? i : (p[i] = findSet(p[i]));
		}

		public boolean isSameSet(int i, int j) {
			return findSet(i) == findSet(j);
		}
		public int chunion(int i,int j, int x2) {
			if (isSameSet(i, j))
				return 0;
			numSets--;
			int x = findSet(i), y = findSet(j);
			int z=findSet(x2);
			p[x]=z;;
			p[y]=z;
		return x;
		}

		public void unionSet(int i, int j) {
			if (isSameSet(i, j))
				return;
			numSets--;
			int x = findSet(i), y = findSet(j);
			if (rank[x] > rank[y]) {
				p[y] = x;
				setSize[x] += setSize[y];


			} else {
				p[x] = y;
				setSize[y] += setSize[x];
				if (rank[x] == rank[y])
					rank[y]++;


			}
		}

		

		public int numDisjointSets() {
			return numSets;
		}

		public int sizeOfSet(int i) {
			return setSize[findSet(i)];
		}
	}


	static class Quad implements Comparable<Quad> {
		int u;
		int v;
		char state;
		int turns;

		public Quad(int i, int j, char c, int k) {
			u = i;
			v = j;
			state = c;
			turns = k;
		}

		public int compareTo(Quad e) {
			return (int) (turns - e.turns);
		}

	}

	static long manhatandistance(long x, long x2, long y, long y2) {
		return Math.abs(x - x2) + Math.abs(y - y2);
	}

	static long fib[];

	static long fib(int n) {
		if (n == 1 || n == 0) {
			return 1;
		}
		if (fib[n] != -1) {
			return fib[n];
		} else
			return fib[n] = ((fib(n - 2) % mod + fib(n - 1) % mod) % mod);
	}

	static class Point  implements Comparable<Point>{
		long x, y;

		Point(long counth, long counts) {
			x = counth;
			y = counts;
		}

		@Override
		public int compareTo(Point p )
		{
			return Long.compare(p.y*1l*x, p.x*1l*y);
		}
	}

	static TreeSet<Long> primeFactors(long N) // O(sqrt(N) / ln sqrt(N))
	{
		TreeSet<Long> factors = new TreeSet<Long>(); // take abs(N) in case of -ve integers
		int idx = 0, p = primes.get(idx);

		while (p * p <= N) {
			while (N % p == 0) {
				factors.add((long) p);
				N /= p;
			}
			if (primes.size() > idx + 1)
				p = primes.get(++idx);
			else
				break;
		}

		if (N != 1) // last prime factor may be > sqrt(N)
			factors.add(N); // for integers whose largest prime factor has a power of 1
		return factors;
	}

	static boolean visited[];

	/**
	 * static int bfs(int s) { Queue<Integer> q = new LinkedList<Integer>();
	 * q.add(s); int count=0; int maxcost=0; int dist[]=new int[n]; dist[s]=0;
	 * while(!q.isEmpty()) {
	 * 
	 * int u = q.remove(); if(dist[u]==k) { break; } for(Pair v: adj[u]) {
	 * maxcost=Math.max(maxcost, v.cost);
	 * 
	 * 
	 * 
	 * if(!visited[v.v]) {
	 * 
	 * visited[v.v]=true; q.add(v.v); dist[v.v]=dist[u]+1; maxcost=Math.max(maxcost,
	 * v.cost); } }
	 * 
	 * } return maxcost; }
	 **/

	static boolean[] vis2;

	static boolean f2 = false;

	static long[][] matMul(long[][] a2, long[][] b, int p, int q, int r) // C(p x r) = A(p x q) x (q x r) -- O(p x q x
																			// r)
	{
		long[][] C = new long[p][r];
		for (int i = 0; i < p; ++i) {
			for (int j = 0; j < r; ++j) {
				for (int k = 0; k < q; ++k) {
					C[i][j] = (C[i][j] + (a2[i][k] % mod * b[k][j] % mod)) % mod;
					C[i][j] %= mod;
				}

			}
		}
		return C;
	}
	public static int[] schuffle(int[] a2) {
		for (int i = 0; i < a2.length; i++) {
			int x =  (int) (Math.random() * a2.length);
			int temp = a2[x];
			a2[x] = a2[i];
			a2[i] = temp;
		}
		return a2;
	}


	static boolean vis[];
	static HashSet<Integer> set = new HashSet<Integer>();

	static long modPow(long ways, long count, long mod) // O(log e)
	{
		ways %= mod;
		long res = 1;
		while (count > 0) {
			if ((count & 1) == 1)
				res = (res * ways) % mod;
			ways = (ways * ways) % mod;
			count >>= 1;
		}
		return res % mod;
	}

	static long gcd(long l, long o) {
		if (o == 0) {
			return l;
		}
		return gcd(o, l % o);
	}

	static int[] isComposite;
	static int[] valid;

	static ArrayList<Integer> primes;
	static ArrayList<Integer> l1;

	
	static TreeSet<Integer> primus = new TreeSet<Integer>();
	
	static void sieveLinear(int N)
	{
		int[] lp = new int[N + 1];								//lp[i] = least prime divisor of i
		for(int i = 2; i <= N; ++i)
		{
			if(lp[i] == 0)
			{
				primus.add(i);
				lp[i] = i;
			}
			int curLP = lp[i];
			for(int p: primus)
				if(p > curLP || p * i > N)
					break;
				else
					lp[p * i] = i;
		}
	}
	

	public static long[] schuffle(long[] a2) {
		for (int i = 0; i < a2.length; i++) {
			int x =  (int) (Math.random() * a2.length);
			long temp = a2[x];
			a2[x] = a2[i];
			a2[i] = temp;
		}
		return a2;
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

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

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

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

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}

		public int[] nxtArr(int n) throws IOException {
			int[] ans = new int[n];
			for (int i = 0; i < n; i++)
				ans[i] = nextInt();
			return ans;
		}
		
	}
	public static int[] sortarray(int a[]) {
		schuffle(a);
		Arrays.sort(a);
		return a;
		}
		public static long[] sortarray(long a[]) {
			schuffle(a);
			Arrays.sort(a);
			return a;
		}
}