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 TreesAndQueries {
	public static int next(StringTokenizer st) {
		return Integer.parseInt(st.nextToken());
	}
 
	static int N, Q;
	static ArrayList<Integer> adjlist[];
	static int[] lvl;
	static LCA lca;
	public static void main(String[] args) throws IOException{
		// TODO Auto-generated method stub
		BufferedReader sc = new BufferedReader(new InputStreamReader(System.in));
		// TODO Auto-generated method stub
 
		N = Integer.parseInt(sc.readLine());
		adjlist = new ArrayList[N];
		lvl = new int[N];
		Arrays.fill(lvl, -1);
		for(int i = 0; i < N; i++) adjlist[i] = new ArrayList<Integer>();
		StringTokenizer st;
		for(int i = 1; i < N; i++) {
			st = new StringTokenizer(sc.readLine());
			int u = next(st) - 1;
			int v = next(st) - 1;
			adjlist[u].add(v);
			adjlist[v].add(u);
		}
		lvl[0] = 0;
		dfs0(0);
		lca = new LCA(adjlist, N, 0);
		Q = Integer.parseInt(sc.readLine());
		for(int i = 0; i < Q; i++) {
			st = new StringTokenizer(sc.readLine());
			int x = next(st) - 1, y = next(st) - 1, a = next(st) - 1, b = next(st) - 1, k = next(st);
			int p0 = dist(a, b);
			int p1 = dist(a, x) + dist(b, y) + 1;
			int p2 = dist(a, y) + dist(b, x) + 1;
			int[] m = new int[2];
			m[0] = m[1] = Integer.MAX_VALUE;
			m[p0 % 2] = Math.min(m[p0 % 2], p0);
			m[p1 % 2] = Math.min(m[p1 % 2], p1);
			m[p2 % 2] = Math.min(m[p2 % 2], p2);
			if(m[k % 2] != Integer.MAX_VALUE) {
				System.out.println(((m[k%2] <= k) ? "YES" : "NO"));
			}
			else {
				int c1 = dist(a, x) + dist(a, y) + 1;
				int c2 = dist(b, x) + dist(b, y) + 1;
				if(k % 2 == 0) {
					if(c1 % 2 == 1) {
						m[0] = m[1] + c1;
					}
					if(c2 % 2 == 1) {
						m[0] = Math.min(m[0], m[1] + c2);
					}
					System.out.println(((m[k%2] <= k) ? "YES" : "NO"));
				}
				else {
					if(c1 % 2 == 1) {
						m[1] = m[0] + c1;
					}
					if(c2 % 2 == 1) {
						m[1] = Math.min(m[1], m[0] + c2);
					}
					System.out.println(((m[k%2] <= k) ? "YES" : "NO"));
				}
			}
		}
		
	}
	
	static int dist(int a, int b) {
		int l = lca.lca(a, b);
		return (lvl[a] + lvl[b] - 2*lvl[l]);
	}
	
	static void dfs0(int curr) {
		for(int c : adjlist[curr]) {
			if(lvl[c] == -1) {
				lvl[c] = lvl[curr] + 1;
				dfs0(c);
			}
		}
	}
 
}
 
class LCA {
	public int[] tour;
	public int[] code;
	public int[] antiCode;
	public int[][] rmq;
	public int[] index;
	public LCA(ArrayList<Integer>[] adjlist, int N, int root) {
		tour = new int[2*N-1];
		rmq = new int[2*N-1][32];
		code = new int[N];
		antiCode = new int[N];
		index = new int[N];
		Arrays.fill(index, -1);
		getCodes(adjlist, root, -1, 0);
		createEulerTour(adjlist, root, -1, tour, 0);
		for(int i = 0; i < tour.length; i++) {
			//System.out.println(tour[i]);
			if(index[tour[i]] == -1) {
				index[tour[i]] = i;
			}
		}
		for(int e = 0; e < tour.length; e++) {
			rmq[e][0] = tour[e];
		}
		for(int i = 1; (1<<i) < 2*N-1; i++) {
			for(int e = 0; e + (1<<i) - 1 < 2*N-1; e++) {
				rmq[e][i] = Math.min(rmq[e][i-1], rmq[e+(1<<(i-1))][i-1]);
			}
		}
	}
	
	public int lca(int a, int b) {
		a = code[a];
		b = code[b];
		if(index[b] < index[a]) {
			int t = a;
			a = b;
			b = t;
		}
		int j = 0;
		while((1<<(j+1)) < index[b]-index[a]+1) j++;
		int coded = Math.min(rmq[index[a]][j], rmq[index[b]-(1<<j)+1][j]);
		return antiCode[coded];
	}
	
	public int getCodes(ArrayList<Integer>[] adjlist, int curr, int p, int currCode) {
		code[curr] = currCode;
		antiCode[currCode] = curr;
		currCode++;
		for(int c : adjlist[curr]) {
			if(c == p) continue;
			currCode = getCodes(adjlist, c, curr, currCode);
		}
		return currCode;
	}
	
	public int createEulerTour(ArrayList<Integer>[] adjlist, int curr, int p, int[] tour, int index) {
		for(int c : adjlist[curr]) {
			if(c == p) continue;
			tour[index] = code[curr];
			index++;
			index = createEulerTour(adjlist, c, curr, tour, index);
		}
		tour[index] = code[curr];
		index++;
		return index;
	}
}