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 
from collections import deque
import sys
input = sys.stdin.readline


class LowestCommonAncestor():
    def __init__(self, tree, root):
        self.n = len(tree)
        self.depth = [0] * self.n
        self.log_size = (self.n).bit_length()
        self.parent = [[-1] * self.n for i in range(self.log_size)]

        q = deque([(root, -1, 0)]) 
        while q:
            v, par, dist = q.pop()
            self.parent[0][v] = par
            self.depth[v] = dist
            for child_v in tree[v]:
                if child_v != par:
                    self.depth[child_v] = dist + 1
                    q.append((child_v, v, dist + 1))

        for k in range(1, self.log_size):
            for v in range(self.n):
                self.parent[k][v] = self.parent[k-1][self.parent[k-1][v]]
            
    def lca(self, u, v):
        if self.depth[u] > self.depth[v]:
            u, v = v, u
        for k in range(self.log_size):
            if (self.depth[v] - self.depth[u] >> k) & 1:
                v = self.parent[k][v]
        if u == v:
            return u
          
        for k in reversed(range(self.log_size)):
            if self.parent[k][u] != self.parent[k][v]:
                u = self.parent[k][u]
                v = self.parent[k][v]
        return self.parent[0][u]


n = int(input())
info = [list(map(int, input().split())) for  i in range(n - 1)]

tree = [[] for i in range(n)]
for i in range(n - 1):
    a, b = info[i]
    a -= 1
    b -= 1
    tree[a].append(b)
    tree[b].append(a)
    
lca = LowestCommonAncestor(tree, 0)

q = int(input())
query = [list(map(int, input().split())) for i in range(q)]
    
for i in range(q):
    x, y, a, b, k = query[i]
    lca_ab = lca.lca(a - 1, b - 1)
    ab = lca.depth[a - 1] + lca.depth[b - 1] - 2 * lca.depth[lca_ab]
    lca_xy = lca.lca(x - 1, y - 1)
    xy = lca.depth[x - 1] + lca.depth[y - 1] - 2 * lca.depth[lca_xy]

    lca_ax = lca.lca(a - 1, x - 1)
    lca_ay = lca.lca(a - 1, y - 1)
    lca_bx = lca.lca(b - 1, x - 1)
    lca_by = lca.lca(b - 1, y - 1)
    
    ax = lca.depth[a - 1] + lca.depth[x - 1] - 2 * lca.depth[lca_ax]
    ay = lca.depth[a - 1] + lca.depth[y - 1] - 2 * lca.depth[lca_ay]
    bx = lca.depth[b - 1] + lca.depth[x - 1] - 2 * lca.depth[lca_bx]
    by = lca.depth[b - 1] + lca.depth[y - 1] - 2 * lca.depth[lca_by]
    if ab <= k and k % 2 == ab % 2:
        print("YES")
    elif ax+1+by <= k and k % 2 == (ax+1+by) % 2:
        print("YES")
    elif ay+1+bx <= k and k % 2 == (ay+1+bx) % 2:
        print("YES")
    else:
        print("NO")