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 Main
{
    static int N , K , ref = 0;
    static int [] head , next , to , level ;
    static int [][] P ;
    static void addEdge(int a , int b , boolean end)
    {
        next[++ref] = head[a] ;
        head[a] = ref ;
        to[ref] = b ;
        if(!end)
            addEdge(b , a , true);
    }
    static void dfs(int u , int p)
    {
        if(p != -1)
            level[u] = level[p] + 1 ;
        P[0][u] = p ;
        for(int i = 1 ; i <= K ; i ++)
            if(P[i - 1][u] != -1)
                P[i][u] = P[i - 1][P[i - 1][u]] ;
        for(int i = head[u] ; i != 0 ; i = next[i])
            if(to[i] != p)
                dfs(to[i] , u);
    }
    static int lca(int u , int v)
    {
        if(level[u] < level[v])return lca(v , u) ;
        for(int d = level[u] - level[v] ; d > 0 ; d &= d - 1)
            u = P[31 - Integer.numberOfLeadingZeros(d & -d)][u];
        if(u == v)return u ;
        for(int i = 31 - Integer.numberOfLeadingZeros(level[v]) ; i >= 0 ; i --)
            if(P[i][u] != -1 && P[i][u] != P[i][v])
            {
                u = P[i][u] ; v = P[i][v] ;
            }
        return P[0][u] ;
    }
    static int dist(int u , int v){return level[u] + level[v] - level[lca(u , v)] * 2;}

    public static void main(String[] args) throws Exception
    {
        Scanner sc = new Scanner();
        PrintWriter out = new PrintWriter(System.out);
        N = sc.nextInt() ;
        K = 31 - Integer.numberOfLeadingZeros(N) ;
        P = new int [K + 1][N] ;
        for(int i = 0 ; i <= K ; i ++)
            Arrays.fill(P[i] , -1);
        head = new int[N] ;
        next = new int [N << 1] ;
        to = new int [N << 1] ;
        level = new int [N] ;
        for(int i = 0 ; i < N - 1 ; i++)
            addEdge(sc.nextInt() - 1 , sc.nextInt() - 1 , false);
        dfs(0 , -1);
        int q = sc.nextInt() ;
        while (q -->0)
        {
            int x = sc.nextInt() - 1, y = sc.nextInt() - 1 ;
            int a = sc.nextInt() - 1 , b = sc.nextInt() - 1 ;
            int k = sc.nextInt() ;
            if((dist(a , b) <= k && Math.abs(k - dist(a , b)) % 2 == 0)
                    ||
                    (dist(a , x) + 1 + dist(y , b) <= k && (dist(a , x) + 1 + dist(y , b) - k) % 2 == 0)
                    ||
                    (dist(a , y) + 1 + dist(x , b) <= k && (dist(a , y) + 1 + dist(x , b) - k) % 2 == 0)
            )
                out.println("YES");
            else
                out.println("NO");
        }
        out.flush();
    }

    static class Scanner
    {
        BufferedReader br;
        StringTokenizer st;

        Scanner() { br = new BufferedReader(new InputStreamReader(System.in)); }

        String next() throws IOException
        {
            while (st == null || !st.hasMoreElements())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

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

}