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 x1304E
   {
   	static ArrayDeque<Integer>[] edges;
      public static void main(String[] args) throws IOException 
      {
   		BufferedReader infile = new BufferedReader(new InputStreamReader(System.in));  
   		StringTokenizer st;
         int T = 1;
         for(int qw=0; qw < T; qw++)
         {
      		st = new StringTokenizer(infile.readLine());
            int N = Integer.parseInt(st.nextToken());
      		edges = new ArrayDeque[N+1];
      		for(int i = 0; i < edges.length; i++)
      			edges[i] = new ArrayDeque<Integer>();
      		for(int a = 1; a < N; a++)
            {
      			st = new StringTokenizer(infile.readLine());
      			int x = Integer.parseInt(st.nextToken());
      			int y = Integer.parseInt(st.nextToken());
      			edges[x].add(y);
      			edges[y].add(x);
            }
      		LCA djj = new LCA(N, edges, 1);
            //new stuff
            int Q = Integer.parseInt(infile.readLine());
            int[] root_distance = new int[N+1];
            dfs(1, root_distance, new boolean[N+1], edges);
            //query
            StringBuilder sb = new StringBuilder();
      		for(int i = 0; i < Q; i++) 
            {
      			st = new StringTokenizer(infile.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());
               String res = "NO";
      			int lca = djj.lca(a, b);
      			int length = root_distance[a]+root_distance[b]-2*root_distance[lca];
               if(length <= k && length%2 == k%2)
                  res = "YES";
               int lca2 = djj.lca(a,x);
               int len2a = root_distance[a]+root_distance[x]-2*root_distance[lca2];
               lca2 = djj.lca(b,y);
               len2a += root_distance[b]+root_distance[y]-2*root_distance[lca2]+1;
               if(len2a <= k && len2a%2 == k%2)
                  res = "YES";
               lca2 = djj.lca(a,y);
               len2a = root_distance[a]+root_distance[y]-2*root_distance[lca2];
               lca2 = djj.lca(b,x);
               len2a += root_distance[b]+root_distance[x]-2*root_distance[lca2]+1;
               if(len2a <= k && len2a%2 == k%2)
                  res = "YES";
               sb.append(res+"\n");
      		}
            System.out.print(sb.toString());
      		//end
         }
   	}
      public static int dfs(int curr, int[] dist, boolean[] seen, ArrayDeque<Integer>[] edges)
      {
         if(seen[curr])
            return 0;
         seen[curr] = true;
         for(int next: edges[curr])
            if(!seen[next])
            {
               dist[next] = dist[curr]+1;
               dfs(next, dist, seen, edges);
            }
         return dist[curr];
      }
   }
   class LCA
   {
      public int N, root;
      public ArrayDeque<Integer>[] edges;
      private int[] enter;
      private int[] exit;
      private int LOG = 17; //change this
      private int[][] dp;
      
      public LCA(int n, ArrayDeque<Integer>[] edges, int r)
      {
         N = n;   root = r;
         enter = new int[N+1];
         exit = new int[N+1];
         dp = new int[N+1][LOG];
         this.edges = edges;
         int[] time = new int[1];
         dfs(root, 0, time);
         dp[1][0] = 1;
         for(int b=1; b < LOG; b++)
            for(int v=1; v <= N; v++)
               dp[v][b] = dp[dp[v][b-1]][b-1];
      }
      private void dfs(int curr, int par, int[] time)
      {
         dp[curr][0] = par;
         enter[curr] = ++time[0];
         for(int next: edges[curr])
            if(next != par)
               dfs(next, curr, time);
         exit[curr] = ++time[0];
      }
      public int lca(int x, int y)
      {
         if(isAnc(x, y))
            return x;
         if(isAnc(y, x))
            return y;
         int curr = x;
         for(int b=LOG-1; b >= 0; b--)
         {
            int temp = dp[curr][b];
            if(!isAnc(temp, y))
               curr = temp;
         }
         return dp[curr][0];
      }
      private boolean isAnc(int anc, int curr)
      {
         return enter[anc] <= enter[curr] && exit[anc] >= exit[curr];
      }
   }