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 sys import stdin,stderr
def rl():
    return [int(w) for w in stdin.readline().split()]

from collections import namedtuple

Path = namedtuple('Path', 'length midlength middle left right')

n, = rl()
adj = [[] for _ in range(n+1)]
paths = [None for _ in range(n+1)]
for _ in range(n-1):
    u,v = rl()
    adj[u].append(v)
    adj[v].append(u)

def append(path, x):
    if not path:
        return Path(1, 1, x, None, None)
    l = path.length + 1
    if l >= 2 * path.midlength:
        return Path(l, path.midlength * 2, x, path, None)
    else:
        return Path(l, path.midlength, path.middle, path.left, append(path.right, x))

queue = [(1,0)]
for i in range(n):
    v, parent = queue[i]
    paths[v] = append(paths[parent], v)
    for u in adj[v]:
        if u != parent:
            queue.append((u,v))

def gcp(x, y):
    if not x or not y:
        return 0
    while x.midlength > y.midlength:
        x = x.left
    while y.midlength > x.midlength:
        y = y.left
    if x.middle == y.middle:
        return x.midlength + gcp(x.right, y.right)
    else:
        return gcp(x.left, y.left)

def dist(x, y):
    px = paths[x]
    py = paths[y]
    return px.length + py.length - 2 * gcp(px, py)

q, = rl()
for _ in range(q):
    x,y,a,b,k = rl()
    ds = [
        dist(a,b),
        dist(a,x) + 1 + dist(y,b),
        dist(a,y) + 1 + dist(x,b),
    ]
    for d in ds:
        if k >= d and (k - d) % 2 == 0:
            print("YES")
            break
    else:
        print("NO")
