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 Sol2{
	public static List<Integer> adj[];
	public static int n, D;
	public static int depth[];
	public static int par[][];
	public static void main(String[] args) throws IOException{
		FastScanner sc = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		n =sc.nextInt();
		adj = new ArrayList[n];
		for(int i=0; i<n; i++) {
			adj[i] = new ArrayList<>();
		}
		for(int i=0; i<n-1; i++) {
			int a= sc.nextInt()-1;
			int b= sc.nextInt()-1;
			adj[a].add(b);
			adj[b].add(a);
		}
		D = Integer.numberOfTrailingZeros(Integer.highestOneBit(n));
		par = new int[D+1][n];
		depth = new int[n];
		for(int []i:par) {
			Arrays.fill(i, -1);
		}
	    ArrayDeque<Integer> q = new ArrayDeque<>();
	    boolean[] seen = new boolean[n];
	    q.add(0);
	    seen[0] = true;
	    while (q.size() > 0)
	    {
	       int i = q.poll();
	       for (int j : adj[i]) if (!seen[j])
	       {
	          seen[j] = true;
	          q.add(j);
	          par[0][j] = i;
	          depth[j] = depth[i]+1;
	       }
	    }

	    for (int d=1; d<=D; d++)
	    {
	       for (int i=0; i<n; i++)
	       {
	          int mid = par[d-1][i];
	          if (mid != -1)
	             par[d][i] = par[d-1][mid];
	       }
	    }
	    int query = sc.nextInt();
	    for(int i=0; i<query; i++) {
	    	int x = sc.nextInt()-1;
	    	int y = sc.nextInt()-1;
	    	int a = sc.nextInt()-1;
	    	int b = sc.nextInt()-1;
	    	int k = sc.nextInt();
	    	boolean good = false;
	    	int one = dist(a,b);
	    	int two = Math.min(dist(a,x)+dist(b,y)+1, dist(a,y)+dist(b,x)+1);
	    	if(one<=k) {
	    		if(one%2==k%2)good = true;
	    	}
	    	if(two<=k) {
	    		if(two%2==k%2)good = true;
	    	}
	    	//System.out.println(query);
	    	if(good)System.out.println("YES");
	    	else System.out.println("NO");
	    }
		out.close();
	}
	public static int walk(int i, int k)
	{
      for (int d=0; d<=D && i != -1; d++)
         if (((1<<d) & k) > 0)
            i = par[d][i];
      return i;
	}	
	public static int lca(int i, int j)
	{
		if (depth[i] > depth[j])
			i = walk(i, depth[i]-depth[j]);
	    if (depth[j] > depth[i])
	        j = walk(j, depth[j]-depth[i]);
	    if (i == j)
	        return i;

	    for (int d=D; d>=0; d--)
	    {
	        if (par[d][i] != par[d][j])
	        {
	        	i = par[d][i];
	            j = par[d][j];
	        }
	    }

	    return par[0][i];
	}
	public static int dist(int i, int j) {
		if(i==j) return 0;
		return depth[i]+depth[j]-2*depth[lca(i,j)];
	}
	public static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner(String s) {
			try {
				br = new BufferedReader(new FileReader(s));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}

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

		String nextToken() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			}
          return st.nextToken();
		}

		String nextLine() {
			st = null;
			try {
				return br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			return null;
		}

		int nextInt() {
			return Integer.parseInt(nextToken());
		}

		long nextLong() {
			return Long.parseLong(nextToken());
		}

		double nextDouble() {
			return Double.parseDouble(nextToken());
		}
	}
}

