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


class RangeMinQuery:
    # O(nlog(n)) construction/space, O(1) range minimum query
    def __init__(self, data):
        self._data = [list(data)]
        n = len(self._data[0])

        # self._data[i][j] stores the min of the segment [j, j + 2 ** i] where i is in [1,2,4,8,...,log(N)]
        w = 1
        while 2 * w <= n:
            prev = self._data[-1]
            self._data.append([min(prev[j], prev[j + w]) for j in range(n - 2 * w + 1)])
            w *= 2

    def query(self, begin, end):
        # Take the min of the overlapping intervals [begin, begin + 2**depth] and [end - 2**depth, end]
        depth = (end - begin).bit_length() - 1
        return min(self._data[depth][begin], self._data[depth][end - (1 << depth)])


class LCA:
    def __init__(self, graph, root=1):
        # Euler tour representation recording every visit to each node in the DFS
        self.eulerTour = []
        # Whether a node is visited and if so, their first index in the euler tour
        self.time = [None] * len(graph)
        # Depth of each node to the root
        self.depth = [None] * len(graph)
        self.depth[root] = 0

        # DFS
        stack = [root]
        while stack:
            node = stack.pop()
            self.eulerTour.append(node)
            if self.time[node] is None:
                self.time[node] = len(self.eulerTour) - 1
                for nbr in graph[node]:
                    if self.time[nbr] is None:
                        self.depth[nbr] = self.depth[node] + 1
                        stack.append(node)
                        stack.append(nbr)

        self.rmq = RangeMinQuery(self.time[node] for node in self.eulerTour)

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

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

    # TODO isAncestor using euler tour?


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])

    tree = LCA(graph)

    ans = []
    for x, y, a, b, k in queries:
        d_ab = tree.dist(a, b)
        d_axyb = tree.dist(a, x) + 1 + tree.dist(y, b)
        d_ayxb = tree.dist(a, y) + 1 + tree.dist(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__":
    input = sys.stdin.readline
    N = int(input())
    edges = []
    for i in range(N - 1):
        edges.append(list(map(int, input().split())))
    Q = int(input())
    queries = []
    for i in range(Q):
        queries.append(list(map(int, input().split())))
    ans = solve(N, edges, queries)
    print(ans)
