Gildong was hiking a mountain, walking by millions of trees. Inspired by them, he suddenly came up with an interesting idea for trees in data structures: What if we add another edge in a tree?

Then he found that such tree-like graphs are called 1-trees. Since Gildong was bored of solving too many tree problems, he wanted to see if similar techniques in trees can be used in 1-trees as well. Instead of solving it by himself, he's going to test you by providing queries on 1-trees.

First, he'll provide you a tree (not 1-tree) with n vertices, then he will ask you q queries. Each query contains 5 integers: x, y, a, b, and k. This means you're asked to determine if there exists a path from vertex a to b that contains exactly k edges after adding a bidirectional edge between vertices x and y. A path can contain the same vertices and same edges multiple times. All queries are independent of each other; i.e. the added edge in a query is removed in the next query.

Input

The first line contains an integer n (3 ≤ n ≤ 10^5), the number of vertices of the tree.

Next n-1 lines contain two integers u and v (1 ≤ u,v ≤ n, u ≠ v) each, which means there is an edge between vertex u and v. All edges are bidirectional and distinct.

Next line contains an integer q (1 ≤ q ≤ 10^5), the number of queries Gildong wants to ask.

Next q lines contain five integers x, y, a, b, and k each (1 ≤ x,y,a,b ≤ n, x ≠ y, 1 ≤ k ≤ 10^9) – the integers explained in the description. It is guaranteed that the edge between x and y does not exist in the original tree.

Output

For each query, print "YES" if there exists a path that contains exactly k edges from vertex a to b after adding an edge between vertices x and y. Otherwise, print "NO".

You can print each letter in any case (upper or lower).

Example

Input


5
1 2
2 3
3 4
4 5
5
1 3 1 2 2
1 4 1 3 2
1 4 1 3 3
4 2 3 3 9
5 2 3 3 9


Output


YES
YES
NO
YES
NO

Note

The image below describes the tree (circles and solid lines) and the added edges for each query (dotted lines).

<image>

Possible paths for the queries with "YES" answers are: 

  * 1-st query: 1 – 3 – 2 
  * 2-nd query: 1 – 2 – 3 
  * 4-th query: 3 – 4 – 2 – 3 – 4 – 2 – 3 – 4 – 2 – 3 
/*
If you want to aim high, aim high
Don't let that studying and grades consume you
Just live life young
*/
import java.io.*;
import java.util.*;

   public class x1304E
   {
   	static int[] p;
   	static int[] depth;
   	static int[][] anc;
   	static int[] amt;
   	static LinkedList<Integer>[] edges;
   	static LinkedList<Integer> revOrder;
      
   	public static void main(String[] args) throws IOException 
      {
   		BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));  
   		StringTokenizer st;
         int T = 1;
         for(int qw=0; qw < T; qw++)
         {
      		st = new StringTokenizer(infile.readLine());
            int N = Integer.parseInt(st.nextToken());
      		p = new int[N+1];
      		amt = new int[N+1];
      		depth = new int[N+1];
      		Arrays.fill(p, -1);
      		p[0] = p[1] = 0;
      		anc = new int[N+1][20];
      		edges = new LinkedList[N+1];
      		revOrder = new LinkedList<Integer>();
      		for(int i = 0; i < edges.length; i++)
      			edges[i] = new LinkedList<Integer>();
      		for(int a = 1; a < N; a++)
            {
      			st = new StringTokenizer(infile.readLine());
      			int x = Integer.parseInt(st.nextToken());
      			int y = Integer.parseInt(st.nextToken());
      			edges[x].add(y);
      			edges[y].add(x);
            }
      		bfs();
      		genLCA();
            //new stuff
            int Q = Integer.parseInt(infile.readLine());
            int[] root_distance = new int[N+1];
            dfs(1, root_distance, new boolean[N+1], edges);
            //query
            StringBuilder sb = new StringBuilder();
      		for(int i = 0; i < Q; i++) 
            {
      			st = new StringTokenizer(infile.readLine());
      			int x = Integer.parseInt(st.nextToken());
               int y = Integer.parseInt(st.nextToken());
      			int a = Integer.parseInt(st.nextToken());
               int b = Integer.parseInt(st.nextToken());
               int k = Integer.parseInt(st.nextToken());
               String res = "NO";
      			int lca = lca(a, b);
      			int length = root_distance[a]+root_distance[b]-2*root_distance[lca];
               if(length <= k && length%2 == k%2)
                  res = "YES";
               int lca2 = lca(a,x);
               int len2a = root_distance[a]+root_distance[x]-2*root_distance[lca2];
               lca2 = lca(b,y);
               len2a += root_distance[b]+root_distance[y]-2*root_distance[lca2]+1;
               if(len2a <= k && len2a%2 == k%2)
                  res = "YES";
               lca2 = lca(a,y);
               len2a = root_distance[a]+root_distance[y]-2*root_distance[lca2];
               lca2 = lca(b,x);
               len2a += root_distance[b]+root_distance[x]-2*root_distance[lca2]+1;
               if(len2a <= k && len2a%2 == k%2)
                  res = "YES";
               sb.append(res+"\n");
      		}
            System.out.print(sb.toString());
      		//end
         }
   	}
      public static int dfs(int curr, int[] dist, boolean[] seen, LinkedList<Integer>[] edges)
      {
         if(seen[curr])
            return 0;
         seen[curr] = true;
         for(int next: edges[curr])
            if(!seen[next])
            {
               dist[next] = dist[curr]+1;
               dfs(next, dist, seen, edges);
            }
         return dist[curr];
      }
   	public static void compute() 
      {
   		while(!revOrder.isEmpty()) 
         {
   			int curr = revOrder.removeFirst();
   			amt[p[curr]] += amt[curr];
   		}
   	}
   	public static int lca(int a, int b) 
      {
   		if(depth[a] > depth[b])
   			return lca(b, a);
   		if(depth[a] < depth[b])
   			b = getP(b, depth[a]);
   		for(int k = 19; k > 0; k--)
   			while(anc[a][k] != anc[b][k]) 
            {
   				a = anc[a][k];
   				b = anc[b][k];
   			}
   		while(a != b) 
         {
   			a = p[a];
   			b = p[b];
   		}
   		return a;
   	}
   	public static int getP(int curr, int wantedD) 
      {
   		for(int k = 19; depth[curr] != wantedD; k--)
   			while(depth[curr] - (1<<k) >= wantedD)
   				curr = anc[curr][k];
   		return curr;
   	}
   	public static void genLCA() 
      {
   		for(int i = 1; i < p.length; i++)
   			anc[i][0] = p[i];
   		for(int j = 1; j < anc[0].length; j++) 
   			for(int i = 1; i < p.length; i++) 
   				anc[i][j] = anc[anc[i][j-1]][j-1];
   	}
   	public static void bfs() 
      {
   		LinkedList<Integer> q = new LinkedList<Integer>();
   		q.add(1);
   		while(!q.isEmpty()) 
         {
   			int curr = q.removeFirst();
   			revOrder.addFirst(curr);
   			for(int child: edges[curr])
   				if(p[child] == -1) 
               {
   					p[child] = curr;
   					depth[child] = 1 + depth[curr];
   					q.add(child);
   				}
   		}
   	}
   }