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 BaseCodeForCoding;

import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class TEST {
	static BufferedReader in;
	static PrintWriter out;

	public static void main(String[] args) throws IOException {
		setIO();
		int N = Integer.parseInt(in.readLine());

		Tree t = new Tree(N);

		for (int i = 0; i < N - 1; i++)
			t.addEdge(in.readLine());

		int Q = Integer.parseInt(in.readLine());

		for (int q = 0; q < Q; q++) {
			String[] tokens = in.readLine().split(" ");
			int x = Integer.parseInt(tokens[0]) - 1;
			int y = Integer.parseInt(tokens[1]) - 1;
			int a = Integer.parseInt(tokens[2]) - 1;
			int b = Integer.parseInt(tokens[3]) - 1;
			int k = Integer.parseInt(tokens[4]);

			int ab = t.dist(a, b);
			int axyb = t.dist(a, x) + 1 + t.dist(y, b);
			int ayxb = t.dist(a, y) + 1 + t.dist(x, b);

			if (check(ab, k) || check(axyb, k) || check(ayxb, k)) {
				System.out.println("YES");
			} else {
				System.out.println("NO");
			}

		}
		out.close();
	}

	public static boolean check(int dist, int k) {
		return dist <= k && (dist % 2 == k % 2);
	}

	static class Tree {
		int N;
		int log;
		List<Integer>[] adj;
		int[][] anc;
		int[] depth;
		int edgeCount;

		Tree(int v) {
			N = v;
			log = log(N);
			adj = new List[v];
			anc = new int[N][log];
			depth = new int[N];
			edgeCount = 0;
			for (int i = 0; i < adj.length; i++) {
				adj[i] = new ArrayList<>();
				Arrays.fill(anc[i], -1);
			}

		}

		void addEdge(int src, int dest) {
			adj[src].add(dest);
			adj[dest].add(src);
			if (++edgeCount == N - 1)
				dfs(0, 0);
		}

		void addEdge(String s) {
			String[] tokens = s.split(" ");
			int u = Integer.parseInt(tokens[0]) - 1;
			int v = Integer.parseInt(tokens[1]) - 1;
			addEdge(u, v);
		}

		void dfs(int u, int p) {
			anc[u][0] = p;
			for (int i = 1; i < log; i++)
				anc[u][i] = anc[anc[u][i - 1]][i - 1];
			for (int v : adj[u]) {
				if (v != p) {
					depth[v] = depth[u] + 1;
					dfs(v, u);
				}
			}
		}

		int lca(int u, int v) {
			if (depth[u] < depth[v]) {
				int temp = u;
				u = v;
				v = temp;
			}
			for (int i = log - 1; i >= 0; i--) {
				if ((depth[u] - (1 << i)) >= depth[v])
					u = anc[u][i];
			}
			if (u == v)
				return u;
			for (int i = log - 1; i >= 0; i--) {
				if (anc[u][i] != anc[v][i]) {
					u = anc[u][i];
					v = anc[v][i];
				}
			}
			return anc[u][0];
		}

		int dist(int u, int v) {
			int lcaDepth = depth[lca(u, v)];
			return depth[u] + depth[v] - (lcaDepth << 1);
		}

		private static int log(int i) {
			return Integer.numberOfTrailingZeros(Integer.highestOneBit((i >> 1 << 1) + 1)) + 1;
		}

	}

	public static void setIO() {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		}
	}

	public static void setIO(String s) {
		try {
			in = new BufferedReader(new FileReader(s + ".in"));
			out = new PrintWriter(new BufferedWriter(new FileWriter(s + ".out")));
		} catch (Exception e) {
			e.printStackTrace();
			System.exit(1);
		}
	}

	public static int max(int... a) {
		int max = a[0];
		for (int i = 1; i < a.length; i++) {
			max = Math.max(max, a[i]);
		}
		return max;
	}
}
