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.InputMismatchException;
import java.io.IOException;
import java.util.ArrayList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author KharYusuf
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        FastReader in = new FastReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        E1TreesAndQueries solver = new E1TreesAndQueries();
        solver.solve(1, in, out);
        out.close();
    }

    static class E1TreesAndQueries {
        public void solve(int testNumber, FastReader s, PrintWriter w) {
            int n = s.nextInt();
            ArrayList<Integer>[] aw = new ArrayList[n];
            for (int i = 0; i < n; i++) aw[i] = new ArrayList<>();
            for (int i = 1; i < n; i++) {
                int u = s.nextInt() - 1, v = s.nextInt() - 1;
                aw[u].add(v);
                aw[v].add(u);
            }
            LCA l = new LCA(n, aw);
            l.build(0);
            int q = s.nextInt();
            while (q-- > 0) {
                int x = s.nextInt() - 1, y = s.nextInt() - 1, a = s.nextInt() - 1, b = s.nextInt() - 1, k = s.nextInt();
                int dis1 = l.query(a, b), dis2 = l.query(a, x) + l.query(b, y) + 1, dis3 = l.query(b, x) + l.query(a, y) + 1;
                //w.println(dis1+" "+dis2+" "+dis3);
                w.println((dis1 <= k && dis1 % 2 == k % 2) || (dis2 <= k && dis2 % 2 == k % 2) || (dis3 <= k && dis3 % 2 == k % 2) ? "YES" : "NO");
            }
        }

        public class LCA {
            int[][] par;
            int[] lvl;
            int n;
            final int LOG;
            ArrayList<Integer>[] a;

            LCA(int n, ArrayList<Integer>[] aa) {
                a = aa;
                this.n = n;
                LOG = log2(n) + 1;
                par = new int[n][LOG];
                lvl = new int[n];
            }

            void build(int u) {
                dfs(u, -1, 0);
                for (int j = 1; j < LOG; j++)
                    for (int i = 0; i < n; i++) {
                        par[i][j] = -1;
                        if (par[i][j - 1] != -1) par[i][j] = par[par[i][j - 1]][j - 1];
                    }
            }

            void dfs(int cur, int p, int l) {
                par[cur][0] = p;
                lvl[cur] = l;
                for (int i : a[cur]) if (i != p) dfs(i, cur, l + 1);
            }

            int query(int u, int v) {
                if (lvl[u] < lvl[v]) {
                    int t = u;
                    u = v;
                    v = t;
                }
                int dis = lvl[u] - lvl[v], tot = dis;
                while (dis > 0) {
                    int raise = log2(dis);
                    u = par[u][raise];
                    dis ^= 1 << raise;
                }
                if (u == v) return tot;
                for (int i = LOG - 1; i >= 0; i--) {
                    if (par[u][i] != -1 && par[u][i] != par[v][i]) {
                        u = par[u][i];
                        v = par[v][i];
                        dis |= 1 << (i + 1);
                    }
                }
                return dis += tot + 2;
            }

            int log2(int i) {
                int cnt = -1;
                while (i > 0) {
                    i >>= 1;
                    cnt++;
                }
                return cnt;
            }

        }

    }

    static class FastReader {
        private InputStream stream;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
        private FastReader.SpaceCharFilter filter;

        public FastReader(InputStream stream) {
            this.stream = stream;
        }

        public int read() {

            if (numChars == -1)
                throw new InputMismatchException();

            if (curChar >= numChars) {

                curChar = 0;

                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }

                if (numChars <= 0)
                    return -1;
            }

            return buf[curChar++];
        }

        public int nextInt() {

            int c = read();

            while (isSpaceChar(c))
                c = read();

            int sgn = 1;

            if (c == '-') {
                sgn = -1;
                c = read();
            }

            int res = 0;

            do {
                if (c < '0' || c > '9')
                    throw new InputMismatchException();

                res *= 10;
                res += c - '0';
                c = read();
            }
            while (!isSpaceChar(c));

            return res * sgn;
        }

        public boolean isSpaceChar(int c) {

            if (filter != null)
                return filter.isSpaceChar(c);

            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public interface SpaceCharFilter {
            public boolean isSpaceChar(int ch);

        }

    }
}

