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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
import java.util.ArrayList;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskE solver = new TaskE();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskE {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int N = in.nextInt();
            Tree tree = new Tree(N);
            for (int i = 0; i < N - 1; i++) {
                int u = in.nextInt() - 1;
                int v = in.nextInt() - 1;
                tree.addEdge(u, v);
            }
            tree.root(0, 0, 0);
            int q = in.nextInt();
            iter:
            while (q-- > 0) {
                int x = in.nextInt() - 1;
                int y = in.nextInt() - 1;
                int a = in.nextInt() - 1;
                int b = in.nextInt() - 1;
                int k = in.nextInt();
                int distab = tree.dist(a, b);
                int distaxb = tree.dist(a, x) + tree.dist(y, b) + 1;
                int distayb = tree.dist(a, y) + tree.dist(x, b) + 1;
                if ((distab % 2 == k % 2 && k >= distab) || (distaxb % 2 == k % 2 && k >= distaxb) || (distayb % 2 == k % 2 && k >= distayb)) {
                    out.println("YES");
                } else {
                    out.println("NO");
                }
            }
        }

        public class Tree {
            Node[] tree;
            int K;
            int[][] lift;
            int[] tin;
            int[] tout;
            int timer = 0;

            public Tree(int N) {
                tree = new Node[N];
                K = (int) Math.ceil(Math.log(N) / Math.log(2)) + 1;
                lift = new int[N][K];
                tin = new int[N];
                tout = new int[N];
                for (int i = 0; i < N; i++) {
                    tree[i] = new Node();
                }
            }

            public void root(int n, int p, int d) {
                tree[n].d = d;
                tree[n].p = p;
                lift[n][0] = p;
                tin[n] = timer++;
                for (int i = 1; i < K; i++) {
                    lift[n][i] = lift[lift[n][i - 1]][i - 1];
                }
                for (int ch : tree[n].ch) {
                    if (ch != p) {
                        root(ch, n, d + 1);
                    }
                }
                tout[n] = timer++;
            }

            int lca(int a, int b) {
                if (is_ancestor(a, b)) {
                    return a;
                }
                if (is_ancestor(b, a)) {
                    return b;
                }
                int u = a;
                for (int i = K - 1; i >= 0; i--) {
                    if (!is_ancestor(lift[u][i], b)) {
                        u = lift[u][i];
                    }
                }
                return lift[u][0];
            }

            boolean is_ancestor(int a, int b) {
                //returns tree on equivalence
                return (tin[a] <= tin[b] && tout[a] >= tout[b]);
            }

            public void addEdge(int a, int b) {
                tree[a].ch.add(b);
                tree[b].ch.add(a);
            }

            int dist(int u, int v) {
                int lca = lca(u, v);
                return tree[u].d + tree[v].d - 2 * tree[lca].d;
            }

            class Node {
                ArrayList<Integer> ch = new ArrayList<>();
                int p;
                int d;

            }

        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

    }
}

