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 
# -*- coding: utf-8 -*-

import math
import collections
import bisect
import heapq
import time
import random
import itertools
import sys
from typing import List
from functools import lru_cache
"""
created by shhuan at 2020/3/18 19:28

"""


def solve(N, edges, queries):
    LIM = 20
    parent = [[0 for _ in range(LIM)] for _ in range(N+1)]
    depth = [0 for _ in range(N + 1)]

    def build(u, p):
        # depth[u] = depth[p] + 1
        # parent[u][0] = p
        # for i in range(1, LIM):
        #     parent[u][i] = parent[parent[u][i-1]][i-1]
        # for v in edges[u]:
        #     if v != p:
        #         build(v, u)
        q = [(u, p)]
        while q:
            nq = []
            for u, p in q:
                depth[u] = depth[p] + 1
                parent[u][0] = p
                for i in range(1, LIM):
                    parent[u][i] = parent[parent[u][i-1]][i-1]
                for v in edges[u]:
                    if v != p:
                        nq.append((v, u))
            q = nq

    def dist(x, y):
        if depth[x] > depth[y]:
            x, y = y, x

        d = 0
        for i in range(LIM-1, -1, -1):
            if depth[parent[y][i]] >= depth[x]:
                y = parent[y][i]
                d += 1 << i

        if x == y:
            return d

        for i in range(LIM-1, -1, -1):
            if parent[x][i] != parent[y][i]:
                x = parent[x][i]
                y = parent[y][i]
                d += 1 << (i + 1)

        return d + 2

    def checkparity(u, v):
        return u <= v and u % 2 == v % 2


    build(1, 0)
    ans = []
    for x, y, a, b, k in quries:
        dxy = 1 if x != y else 0
        d = dist(a, b)
        if checkparity(d, k):
            ans.append('YES')
            continue
        d = dist(a, x) + dist(y, b) + dxy
        if checkparity(d, k):
            ans.append('YES')
            continue
        d = dist(a, y) + dist(x, b) + dxy
        if checkparity(d, k):
            ans.append('YES')
            continue
        ans.append('NO')

    return '\n'.join(ans)



N = int(input())
edges = collections.defaultdict(list)
for i in range(N-1):
    u, v = map(int, input().split())
    edges[u].append(v)
    edges[v].append(u)

M = int(input())
quries = []
for i in range(M):
    x, y, a, b, k = map(int, input().split())
    quries.append((x, y, a, b, k))

print(solve(N, edges, quries))