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

public class OneTrees {
    ArrayList<Integer>[] adj;
    int[][] parentOf;
    int[] level;

    public static void main(String[] args) throws IOException {
        new OneTrees().solve();
    }

    public void solve() throws IOException {
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        int n = Integer.parseInt(f.readLine());
        adj = new ArrayList[n];
        for (int i = 0; i < n; i++) {
            adj[i] = new ArrayList<Integer>();
        }
        for (int i = 0; i < n - 1; i++) {
            StringTokenizer tokenizer = new StringTokenizer(f.readLine());
            int a = Integer.parseInt(tokenizer.nextToken()) - 1;
            int b = Integer.parseInt(tokenizer.nextToken()) - 1;
            adj[a].add(b);
            adj[b].add(a);
        }

        parentOf = new int[n][21];
        level = new int[n];
        level[0] = -1;
        construct(0, 0);


        int q = Integer.parseInt(f.readLine());
        for (int q1 = 0; q1 < q; q1++) {
            StringTokenizer tokenizer = new StringTokenizer(f.readLine());
            int x = Integer.parseInt(tokenizer.nextToken()) - 1;
            int y = Integer.parseInt(tokenizer.nextToken()) - 1;
            int a = Integer.parseInt(tokenizer.nextToken()) - 1;
            int b = Integer.parseInt(tokenizer.nextToken()) - 1;
            int k = Integer.parseInt(tokenizer.nextToken());

            int[] possibleModulo = new int[2];
            Arrays.fill(possibleModulo, Integer.MAX_VALUE);

            int distance1 = distanceBetween(a, b);
            int distance2 = Math.min(distanceBetween(a, x) + distanceBetween(y, b) + 1, distanceBetween(a, y) + distanceBetween(x, b) + 1);

            possibleModulo[distance1 % 2] = Math.min(possibleModulo[distance1 % 2], distance1);
            possibleModulo[distance2 % 2] = Math.min(possibleModulo[distance2 % 2], distance2);
            if (possibleModulo[k % 2] <= k) {
                out.println("Yes");
            } else {
                out.println("No");
            }

        }

        out.close();
    }

    private int distanceBetween(int node1, int node2) {
        int temp, temp2;
        if (level[node1] < level[node2]) {
            temp = node1;
            temp2 = node2;
        } else {
            temp = node2;
            temp2 = node1;
        }
        node1 = temp;
        node2 = temp2;

        int distance = 0;

        if (level[node1] != level[node2]) {
            for (int i = 20; i >= 0; i--) {
                if (level[parentOf[node2][i]] > level[node1]) {
                    node2 = parentOf[node2][i];
                    distance += 1 << i;
                }
            }
            node2 = parentOf[node2][0];
            distance++;
        }

        if (node1 == node2) {
            return distance;
        }

        for (int i = 20; i >= 0; i--) {
            if (parentOf[node1][i] != parentOf[node2][i]) {
                node1 = parentOf[node1][i];
                node2 = parentOf[node2][i];
                distance += 1 << (i + 1);
            }
        }
        distance += 2;
        return distance;
    }

    private void construct(int node, int parent) {
        parentOf[node][0] = parent;
        level[node] = level[parent] + 1;
        for (int i = 1; i <= 20; i++) {
            parentOf[node][i] = parentOf[parentOf[node][i - 1]][i - 1];
        }
        for (int adjacent : adj[node]) {
            if (adjacent == parent) continue;
            construct(adjacent, node);
        }
    }
}
