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 round620;
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;

public class E {
	InputStream is;
	PrintWriter out;
	String INPUT = "";
	
	void solve()
	{
		int n = ni();
		int[] from = new int[n - 1];
		int[] to = new int[n - 1];
		for (int i = 0; i < n - 1; i++) {
			from[i] = ni() - 1;
			to[i] = ni() - 1;
		}
		int[][] g = packU(n, from, to);
		int[][] pars = parents3(g, 0);
		int[] par = pars[0], ord = pars[1], dep = pars[2];
		
		int[][] spar = logstepParents(par);
		
		for(int Q = ni();Q > 0;Q--){
			int x = ni()-1, y = ni()-1;
			int a = ni()-1, b = ni()-1;
			int K = ni();
			int d = d(a, b, spar, dep);
			int d2 = d(a, x, spar, dep) + 1 + d(y, b, spar, dep);
			int d3 = d(a, y, spar, dep) + 1 + d(x, b, spar, dep);
			d2 = Math.min(d2, d3);
			if(d % 2 == d2 % 2){
				if(K % 2 == d % 2 && K >= Math.min(d, d2)){
					out.println("YES");
				}else{
					out.println("NO");
				}
			}else{
				if(K % 2 == d % 2){
					out.println(K >= d ? "YES" : "NO");
				}else{
					out.println(K >= d2 ? "YES" : "NO");
				}
			}
		}
	}
	
	int d(int a, int b, int[][] spar, int[] dep)
	{
		int lca = lca2(a, b, spar, dep);
		return dep[a] + dep[b] - 2*dep[lca];
	}
	
	public static int lca2(int a, int b, int[][] spar, int[] depth) {
		if (depth[a] < depth[b]) {
			b = ancestor(b, depth[b] - depth[a], spar);
		} else if (depth[a] > depth[b]) {
			a = ancestor(a, depth[a] - depth[b], spar);
		}

		if (a == b)
			return a;
		int sa = a, sb = b;
		for (int low = 0, high = depth[a], t = Integer.highestOneBit(high), k = Integer
				.numberOfTrailingZeros(t); t > 0; t >>>= 1, k--) {
			if ((low ^ high) >= t) {
				if (spar[k][sa] != spar[k][sb]) {
					low |= t;
					sa = spar[k][sa];
					sb = spar[k][sb];
				} else {
					high = low | t - 1;
				}
			}
		}
		return spar[0][sa];
	}

	protected static int ancestor(int a, int m, int[][] spar) {
		for (int i = 0; m > 0 && a != -1; m >>>= 1, i++) {
			if ((m & 1) == 1)
				a = spar[i][a];
		}
		return a;
	}

	public static int[][] logstepParents(int[] par) {
		int n = par.length;
		int m = Integer.numberOfTrailingZeros(Integer.highestOneBit(n - 1)) + 1;
		int[][] pars = new int[m][n];
		pars[0] = par;
		for (int j = 1; j < m; j++) {
			for (int i = 0; i < n; i++) {
				pars[j][i] = pars[j - 1][i] == -1 ? -1 : pars[j - 1][pars[j - 1][i]];
			}
		}
		return pars;
	}


	public static int[][] parents3(int[][] g, int root) {
		int n = g.length;
		int[] par = new int[n];
		Arrays.fill(par, -1);

		int[] depth = new int[n];
		depth[0] = 0;

		int[] q = new int[n];
		q[0] = root;
		for (int p = 0, r = 1; p < r; p++) {
			int cur = q[p];
			for (int nex : g[cur]) {
				if (par[cur] != nex) {
					q[r++] = nex;
					par[nex] = cur;
					depth[nex] = depth[cur] + 1;
				}
			}
		}
		return new int[][] { par, q, depth };
	}

	static int[][] packU(int n, int[] from, int[] to) {
		int[][] g = new int[n][];
		int[] p = new int[n];
		for (int f : from)
			p[f]++;
		for (int t : to)
			p[t]++;
		for (int i = 0; i < n; i++)
			g[i] = new int[p[i]];
		for (int i = 0; i < from.length; i++) {
			g[from[i]][--p[from[i]]] = to[i];
			g[to[i]][--p[to[i]]] = from[i];
		}
		return g;
	}



	void run() throws Exception
	{
		is = oj ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);
		
		long s = System.currentTimeMillis();
		solve();
		out.flush();
		tr(System.currentTimeMillis()-s+"ms");
	}
	
	public static void main(String[] args) throws Exception { new E().run(); }
	
	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;
	
	private int readByte()
	{
		if(lenbuf == -1)throw new InputMismatchException();
		if(ptrbuf >= lenbuf){
			ptrbuf = 0;
			try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
			if(lenbuf <= 0)return -1;
		}
		return inbuf[ptrbuf++];
	}
	
	private boolean isSpaceChar(int c) { return !(c >= 33 && c <= 126); }
	private int skip() { int b; while((b = readByte()) != -1 && isSpaceChar(b)); return b; }
	
	private double nd() { return Double.parseDouble(ns()); }
	private char nc() { return (char)skip(); }
	
	private String ns()
	{
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while(!(isSpaceChar(b))){ // when nextLine, (isSpaceChar(b) && b != ' ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}
	
	private char[] ns(int n)
	{
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while(p < n && !(isSpaceChar(b))){
			buf[p++] = (char)b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}
	
	private char[][] nm(int n, int m)
	{
		char[][] map = new char[n][];
		for(int i = 0;i < n;i++)map[i] = ns(m);
		return map;
	}
	
	private int[] na(int n)
	{
		int[] a = new int[n];
		for(int i = 0;i < n;i++)a[i] = ni();
		return a;
	}
	
	private int ni()
	{
		int num = 0, b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private long nl()
	{
		long num = 0;
		int b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}
		
		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}
	
	private boolean oj = System.getProperty("ONLINE_JUDGE") != null;
	private void tr(Object... o) { if(!oj)System.out.println(Arrays.deepToString(o)); }
}
