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 
# by the authority of GOD     author: manhar singh sachdev #

import os,sys
from io import BytesIO, IOBase

def euler_path(n,path):
    height = [0]*n+[10**10]
    euler,st,visi,he = [],[0],[1]+[0]*(n-1),0
    first = [-1]*n
    while len(st):
        x = st[-1]
        euler.append(x)
        if first[x] == -1:
            first[x] = len(euler)-1
        while len(path[x]) and visi[path[x][-1]]:
            path[x].pop()
        if not len(path[x]):
            he -= 1
            st.pop()
        else:
            i = path[x].pop()
            he += 1
            st.append(i)
            height[i],visi[i] = he,1
    return height,euler,first

def cons(euler,height):
    n = len(euler)
    xx = n.bit_length()
    dp = [[n]*n for _ in range(xx)]
    dp[0] = euler
    for i in range(1,xx):
        for j in range(n-(1<<i)+1):
            a,b = dp[i-1][j],dp[i-1][j+(1<<(i-1))]
            dp[i][j] = a if height[a] < height[b] else b
    return dp

def lca(l,r,dp,height,first):
    l,r = first[l],first[r]
    if l > r:
        l,r = r,l
    xx1 = (r-l+1).bit_length()-1
    a,b = dp[xx1][l],dp[xx1][r-(1<<xx1)+1]
    return a if height[a] < height[b] else b

def dis(a,b,dp,height,first):
    return height[a]+height[b]-2*height[lca(a,b,dp,height,first)]

def main():
    n = int(input())
    path = [[] for _ in range(n)]
    for _ in range(n-1):
        u1,v1 = map(int,input().split())
        path[u1-1].append(v1-1)
        path[v1-1].append(u1-1)
    height,euler,first = euler_path(n,path)
    dp = cons(euler,height)
    for _ in range(int(input())):
        x,y,a,b,k = map(lambda xx:int(xx)-1,input().split())
        k += 1
        z = dis(a,b,dp,height,first)
        z1 = dis(a,x,dp,height,first)+1+dis(y,b,dp,height,first)
        z2 = dis(a,y,dp,height,first)+1+dis(x,b,dp,height,first)
        if z <= k and z&1 == k&1:
            print('YES')
        elif z1 <= k and z1&1 == k&1:
            print('YES')
        elif z2 <= k and z2&1 == k&1:
            print('YES')
        else:
            print('NO')

# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0
    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()
    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()
    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
if __name__ == "__main__":
    main()