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.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

public class A {

    static int N;
    static ArrayList<Integer>[] adjList;
    static int[] tree;
    static int[] level;
    static int[][] P;

    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        N = sc.nextInt();
        adjList = new ArrayList[N];
        for (int i = 0; i < N; i++) {
            adjList[i] = new ArrayList<>();
        }
        for (int i = 0; i < N - 1; i++) {
            int u = sc.nextInt() - 1, v = sc.nextInt() - 1;
            adjList[u].add(v);
            adjList[v].add(u);
        }

        tree = new int[N];
        level = new int[N];
        dfs(0, -1, 0);

        constructLca();

        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();
            int originalDist = shortestPath(a, b);
            if (validDist(originalDist, k)) {
                out.println("YES");
            } else {
                int newDist = shortestPath(a, x) + 1 + shortestPath(y, b);
                if (validDist(newDist, k)) {
                    out.println("YES");
                } else {
                    newDist = shortestPath(a, y) + 1 + shortestPath(x, b);
                    if (validDist(newDist, k)) {
                        out.println("YES");
                    } else {
                        out.println("NO");
                    }
                }
            }
        }
        out.flush();
        out.close();
    }

    static boolean validDist(int dist, int k) {
        return k - dist >= 0 && (k - dist) % 2 == 0;
    }

    static void constructLca() {
        int k = 0; while (1<<k <= N) k++;
        P = new int[N][k + 1];
        for (int[] a : P) Arrays.fill(a, -1);
        for (int i = 0; i < N; i++) {
            P[i][0] = tree[i];
        }
        for (int i = 1; i <= k; i++) {
            for (int j = 0; j < N; j++) {
                if (P[j][i - 1] != -1) {
                    P[j][i] = P[P[j][i - 1]][i - 1];
                }
            }
        }
    }

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

    static int lca(int u, int v) {
        // u ta7t --> level[u] akbar
        if (level[u] < level[v]) {
            int tmp = u;
            u = v;
            v = tmp;
        }

        int k = 0; while (level[u] - (1 << (k + 1)) >= 0) k++;

        for (int i = k; i >= 0; i--) {
            if (level[u] - (1 << i) >= level[v]) {
                u = P[u][i];
            }
        }

        if (u == v) return u;

        for (int i = k; i >= 0; i--) {
            if (P[u][i] != -1 && P[u][i] != P[v][i]) {
                u = P[u][i];
                v = P[v][i];
            }
        }
        return P[u][0];
    }

    static void dfs(int u, int p, int l) {
        tree[u] = p;
        level[u] = l;
        for (int v : adjList[u]) {
            if (v != p) {
                dfs(v, u, l + 1);
            }
        }
    }

    static class Scanner
    {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream system) {br = new BufferedReader(new InputStreamReader(system));}
        public Scanner(String file) throws Exception {br = new BufferedReader(new FileReader(file));}
        public String next() throws IOException
        {
            while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }
        public String nextLine()throws IOException{return br.readLine();}
        public int nextInt() throws IOException {return Integer.parseInt(next());}
        public double nextDouble() throws IOException {return Double.parseDouble(next());}
        public char nextChar()throws IOException{return next().charAt(0);}
        public Long nextLong()throws IOException{return Long.parseLong(next());}
        public boolean ready() throws IOException{return br.ready();}
        public void waitForInput() throws InterruptedException {Thread.sleep(3000);}
    }
}
