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, stdout
import math

graph = []
lca = []
depth = []
lim = 0

def build_lca(cur, p):
    global lca
    global depth
    global lim
    global graph
    stack = []
    stack.append((cur,p))
    while stack:
        cur,p = stack.pop()
        lca[cur][0] = p
        depth[cur] = depth[p] + 1
        for i in range(1, lim):
            lca[cur][i] = lca[lca[cur][i-1]][i-1]
            
        for i in graph[cur]:
            if i == p:
                continue
            stack.append((i,cur))
            
def get_lca(a,b):
    global lim
    global graph
    global lca
    global depth
    
    lg = 0
    if depth[a] > depth[b]:
        a,b = b,a
    for i in range(lim,-1,-1):
        if depth[lca[b][i]] >= depth[a]:
            b = lca[b][i]
            lg += int(pow(2,i))

    if b == a:
        return lg
    for i in range(lim,-1,-1):
        if lca[b][i] != lca[a][i]:
            b = lca[b][i]
            a = lca[a][i]
            lg += int(pow(2,i+1))
    return lg +2
    

def main():
    global graph
    global lca
    global depth
    global lim
    n = int(stdin.readline())
    lim =int(math.log(n,2))+1
    graph = [set() for _ in range(n+1)]
    lca = [[0 for _ in range(lim+1)] for _ in range(n+1)]
    depth = [0 for _ in range(n+1)]
    
    for _ in range (n-1):
        a,b = list(map(int, stdin.readline().split()))
        graph[a].add(b)
        graph[b].add(a)
    build_lca(1,0)
    q = int(stdin.readline())
    for _ in range(q):
        x,y,a,b,k = list(map(int, stdin.readline().split()))
        ans = -1
        temp = get_lca(a,b)
        if temp % 2 == k % 2 and k >= temp:
            stdout.write("YES\n")
            continue
        temp = min(get_lca(x,a) + get_lca(y,b), get_lca(x,b) + get_lca(y,a)) + 1
        if temp % 2 == k % 2 and k >= temp:
            stdout.write("YES\n")
            continue
        stdout.write("NO\n")
        
main()
        