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 D {

	static int n, timer, tin[], tout[],level[];
	static int log = 22;
	static int up[][];
	static ArrayList<Integer>[] adj;

	static boolean isAnc(int u, int v) {
		return tin[v] >= tin[u] && tout[v] <= tout[u];
	}

	static int lca(int u, int v) {

		if(isAnc(u, v))return u;
		if(isAnc(v, u))return v;
		for (int i = log - 1; i >= 0; i--) {
			int x = up[u][i];
			if (!isAnc(x, v))
				u = x;
		}

		return up[u][0];
	}

	static int dist(int u, int v) {
		return level[u] + level[v] - 2 * level[lca(u, v)];
	}

	static void dfs(int u, int p) {

		tin[u]=timer++;
		up[u][0] = p;
		for (int i = 1; i < log; i++)
			up[u][i] = up[up[u][i - 1]][i - 1];
		for (int v : adj[u])
			if (v != p) {
				level[v] = level[u] + 1;
				dfs(v, u);
			}
		tout[u]=timer++;

	}

	static boolean can(int from, int to, int k) {
		if (k < 0)
			return false;
		int dis = dist(from, to);
		return dis <= k && dis % 2 == k % 2;
	}

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner();
		PrintWriter out = new PrintWriter(System.out);
		n = sc.nextInt();
		adj = new ArrayList[n];
		for (int i = 0; i < n; i++)
			adj[i] = new ArrayList();
		up = new int[n][log];
		level = new int[n];
		tin=new int [n];
		tout=new int [n];
		for (int i = 1; i < n; i++) {
			int u = sc.nextInt() - 1, v = sc.nextInt() - 1;
			adj[u].add(v);
			adj[v].add(u);
		}
		dfs(0, 0);
		int q = sc.nextInt();
		while (q-- > 0) {
			int x = sc.nextInt() - 1, y = sc.nextInt() - 1, a = sc.nextInt() - 1, b = sc.nextInt() - 1,
					k = sc.nextInt();
			// no edge used
			boolean ans = false;
			ans |= can(a, b, k);
			// go to x then y then b
			int axy = dist(a, x) + 1;
			ans |= can(y, b, k - axy);
			// go to y then x then b
			int ayx = dist(a, y) + 1;
			ans |= can(x, b, k - ayx);
			out.println(ans ? "YES" : "NO");
		}
		out.close();

	}

	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

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

		Scanner(String fileName) throws FileNotFoundException {
			br = new BufferedReader(new FileReader(fileName));
		}

		String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

		String nextLine() throws IOException {
			return br.readLine();
		}

		int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

		long nextLong() throws NumberFormatException, IOException {
			return Long.parseLong(next());
		}

		double nextDouble() throws NumberFormatException, IOException {
			return Double.parseDouble(next());
		}

		boolean ready() throws IOException {
			return br.ready();
		}

	}

	static void sort(int[] a) {
		shuffle(a);
		Arrays.sort(a);
	}

	static void shuffle(int[] a) {
		int n = a.length;
		Random rand = new Random();
		for (int i = 0; i < n; i++) {
			int tmpIdx = rand.nextInt(n);
			int tmp = a[i];
			a[i] = a[tmpIdx];
			a[tmpIdx] = tmp;
		}
	}

}