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 
import sys
input = sys.stdin.readline

N = int(input())
G = [set() for _ in range(N+1)]
for _ in range(N-1):
    a, b = map(int, input().split())
    G[a].add(b)
    G[b].add(a)
S = [None]*(2*N-1)
F = [None]*(N+1)
stack = [1]
visited = set()
depth = [None]*(N+1)
depth[1] = 0
ii = 0
while stack:
    v = stack.pop()
    if v > 0:
        visited.add(v)
        F[v], S[ii] = ii, v
        ii += 1
        for u in G[v]:
            if u in visited:
                continue
            stack += [-v, u]
            depth[u] = depth[v] + 1
    else:
        S[ii] = -v
        ii += 1

INF = (N, None)
M = 2*N; M0 = 2**(M-1).bit_length()
data = [INF]*(2*M0)
for i, v in enumerate(S):
    data[M0-1+i] = (depth[v], i)
for i in range(M0-2, -1, -1):
    data[i] = min(data[2*i+1], data[2*i+2])


def _query(a, b):
    yield INF
    a, b = a + M0, b + M0
    while a < b:
        if b & 1:
            b -= 1
            yield data[b-1]
        if a & 1:
            yield data[a-1]
            a += 1
        a, b = a >> 1, b >> 1


def query(u, v):
    fu, fv = F[u], F[v]
    if fu > fv:
        fu, fv = fv, fu
    return S[min(_query(fu, fv+1))[1]]


def dist(x, y):
    c = query(x, y)
    return depth[x] + depth[y] - 2*depth[c]


Q = int(input())
for _ in range(Q):
    x, y, a, b, k = map(int, input().split())
    t1 = dist(a, b)
    t2 = min(dist(a, x)+dist(y, b), dist(a, y)+dist(x, b)) + 1
    if (t1 <= k and (t1-k) % 2 == 0) or (t2 <= k and (t2-k) % 2 == 0):
        print("YES")
    else:
        print("NO")
