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.io.PrintWriter;
import java.util.Arrays;
import java.util.LinkedList;
public class E{

    static class Edge{
        int to,next;
    }

    static Edge[] edges;
    static int d = 20;
    static int[] head;
    static int tot;
    static int n;
    static void addEdge(int u, int v){

        edges[tot] = new Edge();
        edges[tot].to = v;
        edges[tot].next = head[u];
        head[u] = tot++;
    }


    static void init(){
        tot = 0;
        Arrays.fill(head,-1);
    }

    static int fa[][];

    static int deg[];

    static void bfs(int root){

        LinkedList<Integer> q = new LinkedList<Integer>();
        deg[root] = 0;
        fa[root][0] = 0;
        q.add(root);
        while(!q.isEmpty()){
            int tmp = q.poll();
            for(int i = 1; i < d; ++i){
                fa[tmp][i] = fa[fa[tmp][i-1]][i-1];
            }
            for(int i = head[tmp]; i != -1; i = edges[i].next){
                int v =edges[i].to;
                if(v == fa[tmp][0]){
                    continue;
                }
                deg[v] = deg[tmp] + 1;
                fa[v][0] = tmp;
                q.add(v);
            }
        }
    }

    static int lca(int u, int v){
        if(deg[u] > deg[v]){
            int tp = u;
            u = v;
            v = tp;
        }
        int hu = deg[u];
        int hv = deg[v];
        int tu = u;
        int tv = v;
        for(int det = hv - hu, i = 0; det != 0; det>>=1, ++i){
            if((det & 1) != 0){
                tv = fa[tv][i];
            }
        }
        if(tu == tv){
            return tu;
        }
        for(int i = d-1; i >= 0; --i){
            if(fa[tu][i] == fa[tv][i]){
                continue;
            }
            tu = fa[tu][i];
            tv = fa[tv][i];
        }
        return fa[tu][0];
    }

    static boolean[] flag;

    static Scanner sc = new Scanner(System.in);
    static PrintWriter out = new PrintWriter(System.out);
    public static void main(String[] args){
        n = sc.nextInt();
        flag = new boolean[n+1];
        edges = new Edge[2*n+1];
        deg = new int[n+1];
        head = new int[n+1];
        fa = new int[n+1][d];
        init();
        for(int i = 1; i < n; ++i){
            int u = sc.nextInt();
            int v =sc.nextInt();
            addEdge(u, v);
            addEdge(v, u);
            flag[v] = true;
        }
        int root = 1;
        for(int i = 1; i <= n; ++i){
            if(!flag[i]){
                root = i;
                break;
            }
        }
        bfs(root);
        
        int q = sc.nextInt();
        for(int i = 0; i < q; ++i){

            int x = sc.nextInt();
            int y = sc.nextInt();
            int a = sc.nextInt();
            int b = sc.nextInt();
            int k = sc.nextInt();

            int lcaAB = lca(a, b);
            int lcaAX = lca(a, x);
            int lcaBX = lca(b, x);
            int lcaAY = lca(a, y);
            int lcaBY = lca(b, y);

            int path0 = deg[a] - deg[lcaAB] + deg[b] - deg[lcaAB];
            int path1 = deg[a] - deg[lcaAX] + deg[x] - deg[lcaAX]
                 + deg[b] - deg[lcaBY] + deg[y] - deg[lcaBY] + 1;
            int path2 = deg[a] - deg[lcaAY] + deg[y] - deg[lcaAY]
                 + deg[b] - deg[lcaBX] + deg[x] - deg[lcaBX] + 1;
            if((path0 <= k && (k - path0)%2 == 0) 
                || (path1 <= k && (k - path1)%2 == 0)
                || (path2 <= k && (k - path2)%2 == 0)  ){
                out.println("YES");
            }else{
                out.println("NO");
            }

        }
        out.close();


    }

}
 	  	 			 			 	 	  		 				 	 	