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.io.PrintWriter;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.StringTokenizer;
import java.util.TreeSet;

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
/**
 *
 * @author Andy Phan
 */
public class d {
    static int[][] par;
    static int[] depth;
    public static void main(String[] args) {
        FS in = new FS(System.in);
        PrintWriter out = new PrintWriter(System.out);

        int n = in.nextInt();
        ArrayList<Integer>[] adj = new ArrayList[n];
        for(int i = 0; i < n; i++) adj[i] = new ArrayList<>();
        for(int i = 1; i < n; i++) {
            int u = in.nextInt()-1;
            int v = in.nextInt()-1;
            adj[u].add(v);
            adj[v].add(u);
        }
        par = new int[21][n];
        depth = new int[n];
        ArrayDeque<Integer> q = new ArrayDeque<>();
        q.add(0);
        depth[0] = 1;
        for(int i = 0; i < 21; i++) par[i][0] = -1;
        while(!q.isEmpty()) {
            int next = q.poll();
            for(int i : adj[next]) {
                if(depth[i] == 0) {
                    depth[i] = depth[next]+1;
                    par[0][i] = next;
                    q.add(i);
                }
            }
            for(int i = 1; i < 21; i++) par[i][next] = par[i-1][next] == -1 ? -1 : par[i-1][par[i-1][next]];
        }
        int qu = in.nextInt();
        while(qu-->0) {
            int x = in.nextInt()-1;
            int y = in.nextInt()-1;
            int a = in.nextInt()-1;
            int b = in.nextInt()-1;
            int k = in.nextInt();
            int min = Integer.MAX_VALUE;
            int dist = dist(a, b, 20);
            if(dist%2 == k%2) min = dist;
            dist = Math.min(dist(x, a, 20)+dist(y, b, 20), dist(x, b, 20)+dist(y, a, 20))+1;
            if(dist%2 == k%2) min = Math.min(min, dist);
            if(min <= k) out.println("YES");
            else out.println("NO");
        }
        out.close();
    }
    
    static int dist(int a, int b, int dist) {
        if(a == b) return 0;
        if(depth[a] > depth[b]) {
            int num = Integer.numberOfTrailingZeros(depth[a]-depth[b]);
            return (1<<num)+dist(par[num][a], b, dist);
        }
        if(depth[b] > depth[a]) {
            int num = Integer.numberOfTrailingZeros(depth[b]-depth[a]);
            return (1<<num)+dist(a, par[num][b], dist);
        }
        if(dist == -1) return 2;
        if(par[dist][a] == par[dist][b]) return dist(a, b, dist-1);
        return 2*(1<<dist)+dist(par[dist][a], par[dist][b], dist-1);
    }

    static class FS {

        BufferedReader in;
        StringTokenizer token;

        public FS(InputStream str) {
            in = new BufferedReader(new InputStreamReader(str));
        }

        public String next() {
            if (token == null || !token.hasMoreElements()) {
                try {
                    token = new StringTokenizer(in.readLine());
                } catch (IOException ex) {
                }
                return next();
            }
            return token.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }
        
        public int nextInt(int a) {
            return nextInt();
        }
    }
}
