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


public class  Main{
	static LinkedList<Integer>[]adj;
	static public class LCA {
		
		static int log = 22;
		int[] level, up[],sub;
		LCA(int root,int n) {
			up = new int[n][log];
			level = new int[n];
			sub=new int [n];
			dfs(root, root);
		}
 
		int lca(int u, int v) {
 
			if (level[u] < level[v])
				return lca(v, u);
			for (int i = log - 1; i >= 0; i--) {
				int u2 = up[u][i];
				if (level[u2] >= level[v])
					u = u2;
			}
			if (u == v)
				return u;
			for (int i = log - 1; i >= 0; i--) {
				int u2 = up[u][i], v2 = up[v][i];
				if (u2 != v2) {
					u = u2;
					v = v2;
				}
			}
			return up[u][0];
		}
 
		int dfs(int u, int p) {
 
			int ans=1;
			up[u][0] = p;
			for (int i = 1; i < log; i++)
				up[u][i] = up[up[u][i - 1]][i - 1];
			for (int v : adj[u])
				if (v != p) {
					level[v] = level[u] + 1;
					ans+=dfs(v, u);
				}
			return sub[u]=ans;
 
		}
 
	}
	static int[] dist;
	static void dfs(int i,int par,int d) {
		dist[i]=d;
		for(int j:adj[i]) {
			if(j!=par) {
				dfs(j,i,d+1);
			}
		}
	}
	static LCA l;
	static int distxy(int x,int y) {
		return dist[x]+dist[y]-2*dist[l.lca(x, y)];
	}
	public static void main(String[] args) throws Exception{
		MScanner sc=new MScanner(System.in);
		PrintWriter pw=new PrintWriter(System.out);
		int n=sc.nextInt();
		adj=new LinkedList[n];
		dist=new int[n];
		for(int i=0;i<n;i++)adj[i]=new LinkedList<Integer>();
		for(int i=0;i<n-1;i++) {
			int x=sc.nextInt()-1,y=sc.nextInt()-1;
			adj[x].add(y);
			adj[y].add(x);
		}
		dfs(0, -1, 0);
		l=new LCA(0, n);
		int q=sc.nextInt();
		while(q-->0) {
			int x=sc.nextInt()-1,y=sc.nextInt()-1,a=sc.nextInt()-1,b=sc.nextInt()-1,k=sc.nextInt();
			int distab=distxy(a, b),distax=distxy(a, x),distay=distxy(a, y),distab2,distab3;
			int distby=distxy(b, y);
			distab2=distby+distax+1;
			
			int distbx=distxy(b, x);
			distab3=distbx+distay+1;
			
			if((k%2==distab%2 && distab<=k) || (k%2==distab2%2 && distab2<=k) || (k%2==distab3%2 && distab3<=k)) {
				pw.println("YES");
			}
			else {
				pw.println("NO");
			}
			
		}
		pw.flush();
	}
	static class MScanner {
		StringTokenizer st;
		BufferedReader br;
		public MScanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}
 
		public MScanner(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 int[] takearr(int n) throws IOException {
	        int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public long[] takearrl(int n) throws IOException {
	        long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		public Integer[] takearrobj(int n) throws IOException {
	        Integer[]in=new Integer[n];for(int i=0;i<n;i++)in[i]=nextInt();
	        return in;
		}
		public Long[] takearrlobj(int n) throws IOException {
	        Long[]in=new Long[n];for(int i=0;i<n;i++)in[i]=nextLong();
	        return in;
		}
		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);
		}
	}
}