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

# LCA implementation found from:
# https://github.com/cheran-senthil/PyRival/blob/1971590e96b351288cacb09852896f2383862ab6/pyrival/graphs/lca.py
class RangeQuery:
    def __init__(self, data, func=min):
        self.func = func
        self._data = _data = [list(data)]
        i, n = 1, len(_data[0])
        while 2 * i <= n:
            prev = _data[-1]
            _data.append([func(prev[j], prev[j + i]) for j in range(n - 2 * i + 1)])
            i <<= 1

    def query(self, begin, end):
        depth = (end - begin).bit_length() - 1
        return self.func(
            self._data[depth][begin], self._data[depth][end - (1 << depth)]
        )


class LCA:
    def __init__(self, root, graph):
        self.time = [-1] * len(graph)
        self.path = []
        self.depth = [0] * len(graph)  # MODIFIED
        dfs = [root]
        while dfs:
            node = dfs.pop()
            self.path.append(node)
            if self.time[node] == -1:
                self.time[node] = len(self.path) - 1
                for nei in graph[node]:
                    if self.time[nei] == -1:
                        self.depth[nei] = self.depth[node] + 1  # MODIFIED
                        dfs.append(node)
                        dfs.append(nei)
        self.rmq = RangeQuery(self.time[node] for node in self.path)

    def lca(self, a, b):
        a = self.time[a]
        b = self.time[b]
        if a > b:
            a, b = b, a
        return self.path[self.rmq.query(a, b + 1)]


def solve(N, edges, queries):

    graph = [[] for i in range(N + 1)]
    for e in edges:
        graph[e[0]].append(e[1])
        graph[e[1]].append(e[0])

    L = LCA(1, graph)

    def lca(u, v):
        return L.lca(u, v)

    def getDist(u, v):
        return L.depth[u] + L.depth[v] - 2 * L.depth[lca(u, v)]

    ans = []
    for x, y, a, b, k in queries:
        d_ab = getDist(a, b)
        d_axyb = getDist(a, x) + 1 + getDist(y, b)
        d_ayxb = getDist(a, y) + 1 + getDist(x, b)
        if any(d <= k and (k - d) % 2 == 0 for d in [d_ab, d_axyb, d_ayxb]):
            ans.append("YES")
        else:
            ans.append("NO")
    return "\n".join(ans)


if __name__ == "__main__":
    N = int(input())
    edges = []
    for i in range(N - 1):
        edges.append(list(map(int, sys.stdin.readline().split())))
    Q = int(input())
    queries = []
    for i in range(Q):
        queries.append(list(map(int, sys.stdin.readline().split())))
    ans = solve(N, edges, queries)
    print(ans)
