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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
import java.util.StringTokenizer;

public class TreesNQueries {
	static int maxLogn = 17;
	static List<Set<Integer>> tree;
	static int[][] parentArr;
	static int[] depth;
	
	public static void main(String[] args) {
		InputStream in = System.in;
		InputReader scan = new InputReader(in);
		int n = scan.nextInt();
		depth = new int[n+1];
		parentArr = new int[n+1][maxLogn+1];
		tree = new ArrayList<>();
		for(int i=0;i<=n;i++) {
			tree.add(new HashSet<>());
		}
		for(int i=1;i<n;i++) {
			int v1 = scan.nextInt();
			int v2 = scan.nextInt();
			tree.get(v1).add(v2);
			tree.get(v2).add(v1);
		}
		build(1,0);
		int query = scan.nextInt();
		for(int q=0;q<query;q++) {
			int x = scan.nextInt();
			int y = scan.nextInt();
			int a = scan.nextInt();
			int b = scan.nextInt();
			int k = scan.nextInt();
			int withoutNewPath = getDistance(a,b);
			int withNewPath = Math.min(getDistance(a, x)+getDistance(b, y), getDistance(a, y)+getDistance(x, b))+1;
			int ans = Integer.MAX_VALUE;
			if(withoutNewPath%2 == k%2) {
				ans = withoutNewPath;
			}
			if(withNewPath%2 == k%2) {
				ans = Math.min(ans, withNewPath);
			}
			if(ans<=k) {
				System.out.println("YES");
			} else {
				System.out.println("NO");
			}
		}
	}
	
	static int getDistance(int a,int b) {
		return depth[a] + depth[b] - (2*depth[getLca(a,b)]);
	}
	
	static int getLca(int a,int b) {
		int upper = a;
		int lower = b;
		if(depth[a]>depth[b]) {
			upper = b;
			lower = a;
		}
		for(int i=maxLogn;i>=0;--i) {
			if(depth[parentArr[lower][i]]>=depth[upper]) {
				lower = parentArr[lower][i];
			}
		}
		//Now both lower and upper are at same depth
		if(lower==upper) return lower;
		for(int i=maxLogn;i>=0;--i) {
			if(parentArr[upper][i]!=parentArr[lower][i]) {
				upper = parentArr[upper][i];
				lower = parentArr[lower][i];
			}
		}
		return parentArr[upper][0];
	}
	
	static void build(int current, int parent) {
		depth[current] = depth[parent] + 1;
		parentArr[current][0] = parent;
		for(int i=1;i<=maxLogn;i++) {
			parentArr[current][i] = parentArr[parentArr[current][i-1]][i-1];
		}
		for(int adj: tree.get(current)) {
			if(adj!=parent) {
				build(adj,current);
			}
		}
	}
	
	static class InputReader {
		BufferedReader br;
		StringTokenizer st;
		InputReader(InputStream in) {
			br = new BufferedReader(new InputStreamReader(in));
		}
		
		String next() {
			while(st==null || !st.hasMoreTokens()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
	}
}
