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.util.Scanner;
import java.util.ArrayList;

public class Graph_Tasks {
	private int V;
	private int E;
	private ArrayList<Integer>[] graph;
	private int[][] dp;
	private int[] timeIn;
	private int[] timeOut;
	private int L;
	private int timer = 0;
	
	private Graph_Tasks(int v, int e) {
		V = v + 1;
		graph = new ArrayList[V];
		for (int i = 0; i < V; i++) {
			graph[i] = new ArrayList<Integer>();
		}
		timeIn = new int[V];
		timeOut = new int[V];
		
		L = 1;
		while ((1<<L) <= V) L++;

		dp = new int[V][L + 1];
	}
	
	private void addEdge(int v, int u) {
		graph[v].add(u);
		graph[u].add(v);
	}
	
	private void dfs(int v, int u) {
		timeIn[v] = ++timer;
		dp[v][0] = u;
		
		for (int i = 1; i <= L; i++) {
			dp[v][i] = dp[dp[v][i-1]][i-1];
		}
		
		for (int adj : graph[v]) {
			if (adj != u) {
				dfs(adj, v);
			}
		}
		
		timeOut[v] = ++timer;
	}
	
	private int lca(int v, int u) {
		if (isParent(v, u)) return v;
		if (isParent(u, v)) return u;
		
		for (int i = L; i >= 0; i--) {
			if(!isParent(dp[v][i], u)) {
				v = dp[v][i];
			}
		}
		
		return dp[v][0];
	}
	
	private boolean isParent(int v, int u) {
		return timeIn[v] <= timeIn[u] && timeOut[v]>=timeOut[u];
	}
	
	private String solve(Graph_Tasks g, int x, int y, int a, int b, int k) {
		int a_b_lca = g.lca(a, b);
		int a_x_lca = g.lca(a, x);
		int y_b_lca = g.lca(y, b);
		int a_y_lca = g.lca(a, y);
		int b_x_lca = g.lca(b, x);
		int path_1 = g.pathLength(a, a_b_lca) + g.pathLength(b, a_b_lca);
		int path_2 = g.pathLength(a, a_x_lca) + g.pathLength(x, a_x_lca) + g.pathLength(y, y_b_lca) + g.pathLength(b, y_b_lca) + 1;
		int path_3 = g.pathLength(a, a_y_lca) + g.pathLength(y, a_y_lca) + g.pathLength(b, b_x_lca) + g.pathLength(x, b_x_lca) + 1;
		
		if (k % 2 == 0) {
			if ((path_1 % 2 == 0 && path_1 <= k) || (path_2 % 2 == 0 && path_2 <= k) || (path_3 % 2 == 0 && path_3 <= k)) {
				return "YES";
			}
		} else {
			if ((path_1 % 2 != 0 && path_1 <= k) || (path_2 % 2 != 0 && path_2 <= k) || (path_3 % 2 != 0 && path_3 <= k)) {
				return "YES";
			}
		}
		
		
		return "NO";
	}
	

	private int pathLength(int v, int u) {
		int count = 0;

		while (v != u) {
			if (dp[v][0] == u) {
				count += 1;
				break;
			}
			for (int i = 0; i <= L; i++) {
				if (timeIn[u] >= timeIn[dp[v][i]]) {
					count += Math.pow(2, i-1); 
					v = dp[v][i-1];
					break;
				}
			}
		}
		return count;
	}
	

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		Graph_Tasks g = new Graph_Tasks(n, n-1);

		for (int i = 0; i < n-1; i++) {
			int v = in.nextInt();
			int u = in.nextInt();
			g.addEdge(v, u);
		}
		
		g.dfs(1, 1);
		
		int q = in.nextInt();
		for (int i = 0; i < q; i++) {
			int x = in.nextInt();
			int y = in.nextInt();
			int a = in.nextInt();
			int b = in.nextInt();
			int k = in.nextInt();
			System.out.println(g.solve(g, x, y, a, b, k));
		}
		
		in.close();
	}
}
