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.*;

public class Main {
    public static void main(String[] args){
        InputStream  inputstream;
        OutputStream outputstream;
        inputstream = System.in;
        outputstream = System.out;
        InputReader in = new InputReader(inputstream);
        PrintWriter out = new PrintWriter(outputstream);
        Task solver = new Task();
        solver.solve(in,out);
        out.close();
    }

    static class Task{

        private static final int lg = 18;
        int n,m;
        Vertex[] tree;
        int[][] dp;
        int[] h;

        public void solve(InputReader in,PrintWriter out){
            n = in.nextInt();
            tree = new Vertex[n];
            h = new int[n];
            dp = new int[lg + 2][n];
            Arrays.fill(h,0);
            for(int i = 0;i < n;++i){
                tree[i] = new Vertex();
            }
            for(int i = 0;i < n - 1;++i){
                int u,v;
                u = in.nextInt() - 1;
                v = in.nextInt() - 1;
                tree[u].adj.add(v);
                tree[v].adj.add(u);
            }
            dfs(0,0);
            for(int j = 1;j <= lg;++j){
                for(int i = 0;i < n;++i){
                    dp[j][i] = dp[j - 1][dp[j - 1][i]];
                }
            }
            m = in.nextInt();
            while(m-- > 0){
                int x,y,a,b,k;
                x = in.nextInt() - 1;
                y = in.nextInt() - 1;
                a = in.nextInt() - 1;
                b = in.nextInt() - 1;
                k = in.nextInt();
                boolean ok = false;
                int dist_0 = dist(a,b);
                int dist_1 = dist(a,x) + 1 + dist(y,b);
                int dist_2 = dist(a,y) + 1 + dist(x,b);
                if(((dist_0 & 1) ^ (k & 1)) == 0){
                    ok |= (dist_0 <= k);
                }
                if(((dist_1 & 1) ^ (k & 1)) == 0){
                    ok |= (dist_1 <= k);
                }
                if(((dist_2 & 1) ^ (k & 1)) == 0){
                    ok |= (dist_2 <= k);
                }
                out.print(ok ? "YES\n" : "NO\n");
            }
        }

        void dfs(int u,int par){
            dp[0][u] = par;
            for(int v : tree[u].adj){
                if(v == par){
                    continue;
                }
                h[v] = h[u] + 1;
                dfs(v,u);
            }
        }

        int lca(int u,int v){
            if(h[u] < h[v]){
                int tmp = u;
                u = v;
                v = tmp;
            }
            for(int j = lg;j >= 0;--j){
                if(h[dp[j][u]] >= h[v]){
                    u = dp[j][u];
                }
            }
            if(u == v){
                return u;
            }
            for(int j = lg;j >= 0;--j){
                if(dp[j][u] != dp[j][v]){
                    u = dp[j][u];
                    v = dp[j][v];
                }
            }
            return dp[0][u];
        }

        int dist(int u,int v){
            return h[u] + h[v] - (h[lca(u,v)] << 1);
        }

        public class Vertex{
            ArrayList < Integer > adj;
            Vertex(){
                adj = new ArrayList<>();
            }
        }
    }

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

        public InputReader(InputStream stream){
            reader = new BufferedReader(new InputStreamReader(stream));
            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());
        }

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

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

        public String nextLine(){
            String ret_str = "";
            try {
                ret_str = reader.readLine();
            } catch (IOException e){
                e.printStackTrace();
            }
            return ret_str;
        }
    }
}
