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 E
{
    ArrayList<Integer>[] adj; 
    int n, logn, p[][], h[], enter[], exit[], ind;
    boolean[] vis;

    int lca(int v1, int v2) {
		if(h[v1] < h[v2]) { int temp = v1;  v1 = v2;  v2 = temp; }
		
		for(int i = logn - 1; i >= 0; --i) 
			if(p[v1][i] != -1 && h[p[v1][i]] >= h[v2])
				v1 = p[v1][i];
		
		if(v1 == v2)  return v1;
		for(int i = logn - 1; i >= 0; --i)
			if(p[v1][i] != p[v2][i]) {  v1 = p[v1][i];  v2 = p[v2][i]; }
		return p[v1][0];
    }
    
    void dfs(int node, int par) {
		vis[node] = true;  enter[node] = ind++;
		if(par != -1) {
			p[node][0] = par;  h[node] = h[par] + 1;
			for(int i = 1; i < logn && p[node][i-1] != -1; ++i) 
				p[node][i] = p[p[node][i-1]][i-1];
		} else  h[node] = 0;
		for(int next : adj[node])  if(!vis[next])  dfs(next, node);
		exit[node] = ind;
    }
    
    void solve(FastIO io)
    {
        int n = io.nextInt();
        logn = (int)Math.ceil(Math.log(n)/Math.log(2));
        p = new int[n][logn];  h = new int[n];
        adj = new ArrayList[n];  
        enter = new int[n];  
        exit = new int[n];

        for(int i = 0; i < n; ++i) 
        {
            adj[i] = new ArrayList<Integer>();  
            Arrays.fill(p[i], -1);
		}
		vis = new boolean[n];  ind = 0;

        for (int i = 0; i < n-1; i++)
        {
            int u = io.nextInt() -1;
            int v=  io.nextInt() - 1;

            adj[u].add(v);
            adj[v].add(u);


        }

        dfs(0, -1);

        int q = io.nextInt();
        for (int i = 0; i < q; i++)
        {
            int x = io.nextInt() - 1;
            int y = io.nextInt() - 1;
            int a = io.nextInt() - 1;
            int b = io.nextInt() - 1;
            int k = io.nextInt();

            int min = h[a] + h[b] - 2*h[lca(a, b)];

            int add1 = (h[a] + h[x] - 2*h[lca(a, x)]) + (h[b] + h[y] - 2*h[lca(b, y)]) + 1;
            int add2 = (h[a] + h[y] - 2*h[lca(a, y)]) + (h[b] + h[x] - 2*h[lca(b, x)]) + 1;


            if (min <= k && (k - min) % 2 == 0)
            {
                io.println("YES");
                continue;
            }
            else if (add1 <= k && (k - add1) % 2 == 0)
            {
                io.println("YES");
                continue;
            }
            else if (add2 <= k && (k - add2) % 2 == 0)
            {
                io.println("YES");
                continue;
            }
            else
                io.println("NO");

        }
    }

    public static void main(String[] args)
    {
        FastIO io = new FastIO();
 
        new E().solve(io);

        io.close();
    }

    static class FastIO extends PrintWriter
    {
        BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        FastIO()
        {
            super(System.out);
        }

        public String next()
        {
            while (!st.hasMoreTokens())
            {
                try {
                    st = new StringTokenizer(r.readLine());
                } catch (Exception e) {
                    //TODO: handle exception
                }
            }
            return st.nextToken();
        }

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

        public long nextLong()
        {
            return Long.parseLong(next());
        }

        public double nextDouble()
        {
            return Double.parseDouble(next());
        }
    }
}

