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 
// package Quarantine;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.lang.reflect.Array;
import java.util.*;

public class TreesAndQueries {
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int n=Integer.parseInt(br.readLine());
        StringBuilder print=new StringBuilder();
        ArrayList<Integer> tree[]=new ArrayList[n+1];
        for(int i=1;i<=n;i++){
            tree[i]=new ArrayList<>();
        }
        StringTokenizer st;
        for(int i=1;i<n;i++){
            st=new StringTokenizer(br.readLine());
            int x=Integer.parseInt(st.nextToken());
            int y=Integer.parseInt(st.nextToken());
            tree[x].add(y);
            tree[y].add(x);
        }
        int par[][]=new int[Integer.numberOfTrailingZeros(Integer.highestOneBit(n))+1][n+1];
        for(int a[]:par){
            Arrays.fill(a,-1);
        }
        Queue<Integer> queue=new LinkedList<>();
        boolean visited[]=new boolean[n+1];
        queue.add(1);
        visited[1]=true;
        int depth[]=new int[n+1];
        while(!queue.isEmpty()){
            int curr=queue.remove();
            for(int j:tree[curr]){
                if(!visited[j]){
                    visited[j]=true;
                    queue.add(j);
                    par[0][j]=curr;
                    depth[j]=depth[curr]+1;
                }
            }
        }
        for(int i=1;i<par.length;i++){
            for(int j=1;j<=n;j++){
                int parmid=par[i-1][j];
                if(parmid!=-1){
                    par[i][j]=par[i-1][parmid];
                }
            }
        }
        int q=Integer.parseInt(br.readLine());
        while(q--!=0){
            st=new StringTokenizer(br.readLine());
            int x=Integer.parseInt(st.nextToken());
            int y=Integer.parseInt(st.nextToken());
            int a=Integer.parseInt(st.nextToken());
            int b=Integer.parseInt(st.nextToken());
            int k=Integer.parseInt(st.nextToken());
            int d1=findDistance(a,b,depth,par);
            int d2=findDistance(a,x,depth,par)+1+findDistance(b,y,depth,par);
            int d3=findDistance(a,y,depth,par)+1+findDistance(b,x,depth,par);
            TreeSet<Integer> set=new TreeSet<>();
//            System.out.println(d1+" "+d2+" "+d3);
            if(d1%2==k%2){
                set.add(d1);
            }
            if(d2%2==k%2){
                set.add(d2);
            }
            if(d3%2==k%2){
                set.add(d3);
            }
            if(set.size()==0||set.first()>k){
                print.append("NO\n");
            }
            else{
                print.append("YES\n");
            }
        }
        System.out.print(print.toString());
    }
    public static int travelup(int par[][],int v,int k){
        for(int i=par.length-1;i>=0&&v!=-1;i--){
            if((k&(1<<i))!=0){
                v=par[i][v];
            }
        }
        return v;
    }
    public static int findDistance(int u,int v,int depth[],int par[][]){
        int ans=0;
        if(depth[u]>depth[v]){
            ans+=depth[u]-depth[v];
            u=travelup(par,u,depth[u]-depth[v]);
        }
        if(depth[v]>depth[u]){
            ans+=depth[v]-depth[u];
            v=travelup(par,v,depth[v]-depth[u]);
        }
        if(u==v){
//            System.out.println(ans);
            return ans;
        }
        for(int i=par.length-1;i>=0;i--){
            if(par[i][u]!=par[i][v]){
                u=par[i][u];
                v=par[i][v];
                ans+=2*(1<<i);
            }
        }
        u=par[0][u];
        v=par[0][v];
        ans+=2;
//        System.out.println(ans);
        return ans;
    }
}
