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 
/*
javac e.java && java e
*/

import java.io.*;
import java.util.*;

public class e {
	public static void main(String[] args) { new e(); }
	FS in = new FS();
	PrintWriter out = new PrintWriter(System.out);

	int n, t, lg = 20;
	ArrayList<Integer>[] adj;
	int[] depth;
	int[][] walk;

	e() {
		n = in.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 u = in.nextInt() - 1;
			int v = in.nextInt() - 1;
			adj[u].add(v);
			adj[v].add(u);
		}

		depth = new int[n];
		walk = new int[lg][n];

		int l = 0, r = 0;
		int[] q = new int[3 * n + 1];
		q[r++] = 0; q[r++] = 0;
		while (l != r) {
			int node = q[l++];
			int par = q[l++];
			for (int i : adj[node]) {
				if (i == par) continue;
				walk[0][i] = node;
				depth[i] = 1 + depth[node];
				for (int j = 1; j < lg; j++)
					if (depth[i] >= (1 << j))
						walk[j][i] = walk[j - 1][walk[j - 1][i]];
				q[r++] = i; q[r++] = node;
			}
		}
		
		t = in.nextInt();
		int x, y, a, b, k;
		while (t-- > 0) {
			x = in.nextInt() - 1;
			y = in.nextInt() - 1;
			a = in.nextInt() - 1;
			b = in.nextInt() - 1;
			k = in.nextInt();

			solve(x, y, a, b, k);
		}

		out.close();
	}

	void solve(int x, int y, int a, int b, int k) {
		int p1 = dist(a, b);
		int p2 = dist(a, x) + dist(y, b) + 1;
		int p3 = dist(a, y) + dist(x, b) + 1;
		
		boolean yes = false;
		if (p1 <= k) yes |= ((k - p1) & 1) == 0;
		if (p2 <= k) yes |= ((k - p2) & 1) == 0;
		if (p3 <= k) yes |= ((k - p3) & 1) == 0;

		if (yes) out.println("YES");
		else out.println("NO");
	}

	int dist(int x, int y) {
		return depth[x] + depth[y] - 2 * depth[lca(x, y)];
	}

	int lca(int x, int y) {
		if (x == y) return x;
		if (depth[x] < depth[y])
			x = y^(x^(y = x));
		int del = depth[x] - depth[y];
		for (int i = lg - 1; i >= 0; i--)
			if ((del | (1 << i)) == del)
				x = walk[i][x];
		if (x == y) return x;
		for (int i = lg - 1; i >= 0; i--)
			if (walk[i][x] != walk[i][y]) {
				x = walk[i][x];
				y = walk[i][y];
			}
		return walk[0][x];
	}
	
	int abs(int x) { return x < 0 ? -x : x; }
	long abs(long x) { return x < 0 ? -x : x; }
	double abs(double x) { return x < 0 ? -x : x; }
	int min(int x, int y) { return x < y ? x : y; }	
	int max(int x, int y) { return x > y ? x : y; }
	long min(long x, long y) { return x < y ? x : y; }	
	long max(long x, long y) { return x > y ? x : y; }
	double min(double x, double y) { return x < y ? x : y; }
	double max(double x, double y) { return x > y ? x : y; }
	int gcd(int x, int y) { while (y > 0) { x = y^(x^(y = x)); y %= x; } return x; }
	long gcd(long x, long y) { while (y > 0) { x = y^(x^(y = x)); y %= x; } return x; }
	long lcm(int x, int y) { return ((long) x / gcd(x, y)) * y; }
	long lcm(long x, long y) { return (x / gcd(x, y)) * y; }

	void sort(int[] arr) {
		int sz = arr.length, j;
		Random r = new Random();
		for (int i = 0; i < sz; i++) {
			j = r.nextInt(i + 1);
			arr[i] = arr[j]^(arr[i]^(arr[j] = arr[i]));
		} Arrays.sort(arr);
	}

	class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens()) {
				try { st = new StringTokenizer(br.readLine()); }
				catch (Exception e) {}
			} return st.nextToken();
		}
		int nextInt() { return Integer.parseInt(next()); }
		long nextLong() { return Long.parseLong(next()); }
		double nextDouble() { return Double.parseDouble(next()); }
	}
}
