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.math.BigDecimal;
import java.math.RoundingMode;
import java.util.*;

/*
Прокрастинирую
*/

public class Main {

    static FastReader in;
    static PrintWriter out;
    static Random rand = new Random();
    static final int INF = (int) (1e9 + 10);
    static final int MOD = (int) (1e9 + 7);
    static final int N = (int) (1e6);
    static final int LOGN = 20;
    static ArrayList<Integer>[] g;
    static int[][] p;
    static int[] depth, tin, tout;
    static int timer = 0;

    static void dfs(int v, int from) {
        tin[v] = ++timer;
        p[v][0] = from;
        for (int i = 1; i < LOGN; i++) {
            p[v][i] = p[p[v][i - 1]][i - 1];
        }
        for (int u : g[v]) {
            if (u != from) {
                depth[u] = depth[v] + 1;
                dfs(u, v);
            }
        }
        tout[v] = ++timer;
    }

    static boolean isAnc(int v, int u) {
        return tin[v] < tin[u] && tout[u] < tout[v];
    }
    static int lca(int v, int u) {
        if (v == u) return v;
        if (isAnc(v, u)) return v;
        if (isAnc(u, v)) return u;
        for (int i = LOGN - 1; i >= 0; i--) {
            if (!isAnc(p[v][i], u)) {
                v = p[v][i];
            }
        }
        return p[v][0];
    }

    static int getMinPath(int v, int u) {
        if (v == u) return 0;
        if (isAnc(v, u)) return depth[u] - depth[v];
        if (isAnc(u, v)) return depth[v] - depth[u];
        int vu = lca(v, u);
        return depth[v] + depth[u] - depth[vu] * 2;
    }

    static void solve() {
        int n = in.nextInt();

        g = new ArrayList[n];
        Arrays.setAll(g, gi -> new ArrayList<>());
        for (int i = 0; i < n - 1; i++) {
            int v = in.nextInt() - 1;
            int u = in.nextInt() - 1;
            g[v].add(u);
            g[u].add(v);
        }

        p = new int[n][LOGN];
        depth = new int[n];
        tin = new int[n];
        tout = new int[n];
        dfs(0, 0);

        int q = in.nextInt();
        for (int i = 0; i < q; i++) {
            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 min1 = getMinPath(a, b);
            int min2 = Math.min(getMinPath(a, x) + 1 + getMinPath(y, b), 
                    getMinPath(a, y) + 1 + getMinPath(x, b));

            if ((min1 <= k && (k - min1) % 2 == 0) ||
                    (min2 <= k && (k - min2) % 2 == 0)) {
                out.println("YES");
            } else {
                out.println("NO");
            }
        }
    }

    public static void main(String[] args) throws FileNotFoundException {
        in = new FastReader(System.in);
//        in = new FastReader(new FileInputStream("input.txt"));
        out = new PrintWriter(System.out);
//        out = new PrintWriter(new FileOutputStream("output.txt"));


        int q = 1;
//        q = in.nextInt();

        while (q-- > 0) {
            solve();
        }

        out.close();
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        FastReader(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

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

        Long nextLong() {
            return Long.parseLong(next());
        }

        Double nextDouble() {
            return Double.parseDouble(next());
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(nextLine());
            }
            return st.nextToken();
        }

        String nextLine() {
            String s = "";
            try {
                s = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return s;
        }
    }
}