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.ByteArrayInputStream;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.nio.charset.StandardCharsets;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.InputMismatchException;

public class e {
	
	@SuppressWarnings("unchecked")
	static class Solver {
		
		int N, Q;
		Tree T;
		
		void solve(FastScanner s, PrintWriter out) {
			N = s.nextInt(); T = new Tree(N);
			for(int i = 1, u, v; i < N; i++) {
				u = s.nextInt() - 1; v = s.nextInt() - 1;
				T.add(u, v);
			}
			T.dfs(0, -1);
			Q = s.nextInt();
			while(Q --> 0) {
				int x = s.nextInt() - 1, y = s.nextInt() - 1, a = s.nextInt() - 1, b = s.nextInt() - 1, k = s.nextInt();
				int[] dts = { dist(a, b), dist(a, x) + 1 + dist(y, b), dist(a, y) + 1 + dist(x, b) };
				boolean can = false;
				for(int d : dts) {
					if(k < d) continue;
					if(k % 2 == d % 2) can = true;
				}
				out.println(can ? "YES" : "NO");
			}
		}
		
		int dist(int u, int v) {
			int l = T.lca(u, v);
			return T.h[u] + T.h[v] - 2 * T.h[l];
		}
		
		class Tree {
			ArrayList<Integer>[] adj; int n, logn, p[][], h[], enter[], exit[], ind;
			boolean[] vis;
			
			Tree(int n) {
				this.n = n;  logn = (int)Math.ceil(Math.log(n)/Math.log(2));
				p = new int[n][logn];  h = new int[n];
				adj = new ArrayList[n];  enter = new int[n];  exit = new int[n];
				for(int i = 0; i < n; ++i) {
					adj[i] = new ArrayList<Integer>();  Arrays.fill(p[i], -1);
				}
				
				vis = new boolean[n];  ind = 0;
			}
			
			void add(int v1, int v2) {  adj[v1].add(v2);  adj[v2].add(v1);  }
			
			void dfs(int node, int par) {
				vis[node] = true;  enter[node] = ind++;
				if(par != -1) {
					p[node][0] = par;  h[node] = h[par] + 1;
					for(int i = 1; i < logn && p[node][i-1] != -1; ++i) 
						p[node][i] = p[p[node][i-1]][i-1];
				} else  h[node] = 0;
				for(int next : adj[node])  if(!vis[next])  dfs(next, node);
				exit[node] = ind;
			}

			int lca(int v1, int v2) {
				if(h[v1] < h[v2]) { int temp = v1;  v1 = v2;  v2 = temp; }
				
				for(int i = logn - 1; i >= 0; --i) 
					if(p[v1][i] != -1 && h[p[v1][i]] >= h[v2])
						v1 = p[v1][i];
				
				if(v1 == v2)  return v1;
				for(int i = logn - 1; i >= 0; --i)
					if(p[v1][i] != p[v2][i]) {  v1 = p[v1][i];  v2 = p[v2][i]; }
				return p[v1][0];
			}
		}
		
	}
	
	public static void main(String[] args) {
		FastScanner s = new FastScanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		new Solver().solve(s, out);
		out.close(); s.close();
	}
	
	static class FastScanner {
		private InputStream stream;
		private byte[] buf = new byte[1 << 22];
		private int curChar;
		private int numChars;
		
		public FastScanner(InputStream stream) {
			this.stream = stream;
		}
		
		public FastScanner(File f) throws FileNotFoundException {
			this(new FileInputStream(f));
		}
		
		public FastScanner(String s) {
			this.stream = new ByteArrayInputStream(s.getBytes(StandardCharsets.UTF_8));
		}
		
		void close() {
			try {
				stream.close();
			} catch (IOException e) {
				e.printStackTrace();
			}
		}
		
		int read() {
			if (numChars == -1)
				throw new InputMismatchException();
			if (curChar >= numChars) {
				curChar = 0;
				try {
					numChars = stream.read(buf);
				} catch (IOException e) {
					throw new InputMismatchException();
				}
				if (numChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}
		
		boolean isSpaceChar(int c) {
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}
		
		boolean isEndline(int c) {
			return c == '\n' || c == '\r' || c == -1;
		}
		
		public int nextInt() {
			return Integer.parseInt(next());
		}
		
		public long nextLong() {
			return Long.parseLong(next());
		}
		
		public double nextDouble() {
			return Double.parseDouble(next());
		}
		
		public String next() {
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}
		
		public String nextLine() {
			int c = read();
			while (isEndline(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do {
				res.appendCodePoint(c);
				c = read();
			} while (!isEndline(c));
			return res.toString();
		}
		
		public int[] nextIntArray(int N) {
			int[] ret = new int[N];
			for (int i = 0; i < N; i++)
				ret[i] = this.nextInt();
			return ret;
		}
		
		public int[][] next2DIntArray(int N, int M) {
			int[][] ret = new int[N][];
			for (int i = 0; i < N; i++)
				ret[i] = this.nextIntArray(M);
			return ret;
		}
		
		public long[] nextLongArray(int N) {
			long[] ret = new long[N];
			for (int i = 0; i < N; i++)
				ret[i] = this.nextLong();
			return ret;
		}
		
		public long[][] next2DLongArray(int N, int M) {
			long[][] ret = new long[N][];
			for (int i = 0; i < N; i++)
				ret[i] = nextLongArray(M);
			return ret;
		}
		
		public double[] nextDoubleArray(int N) {
			double[] ret = new double[N];
			for (int i = 0; i < N; i++)
				ret[i] = this.nextDouble();
			return ret;
		}
		
		public double[][] next2DDoubleArray(int N, int M) {
			double[][] ret = new double[N][];
			for (int i = 0; i < N; i++)
				ret[i] = this.nextDoubleArray(M);
			return ret;
		}
		
	}
	
}
