1304_E. 1-Trees and Queries
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 
```python
import sys, os
 
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 = [-1] * len(graph)
        P = [-1] * len(graph)
        t = -1
        dfs = [root]
        while dfs:
            node = dfs.pop()
            self.path[t] = P[node]
            self.time[node] = t = t + 1
            for nei in graph[node]:
                if self.time[nei] == -1:
                    P[nei] = node
                    dfs.append(nei)
        self.rmq = RangeQuery(self.time[node] for node in self.path)
 
    def __call__(self, a, b):
        if a == b:
            return a
        a = self.time[a]
        b = self.time[b]
        if a > b:
            a, b = b, a
        return self.path[self.rmq.query(a, b)]
         
inp = [int(x) for x in sys.stdin.buffer.read().split()]; ii = 0
 
n = inp[ii]; ii += 1
coupl = [[] for _ in range(n)]
for _ in range(n - 1):
    u = inp[ii] - 1; ii += 1
    v = inp[ii] - 1; ii += 1
    coupl[u].append(v)
    coupl[v].append(u)
 
root = 0
lca = LCA(root, coupl)
depth = [-1]*n
depth[root] = 0
bfs = [root]
for node in bfs:
    for nei in coupl[node]: 
        if depth[nei] == -1:
            depth[nei] = depth[node] + 1
            bfs.append(nei)
 
def dist(a,b):
    c = lca(a,b)
    return depth[a] + depth[b] - 2 * depth[c]
 
q = inp[ii]; ii += 1
out = []
for _ in range(q):
    x = inp[ii] - 1; ii += 1
    y = inp[ii] - 1; ii += 1
    a = inp[ii] - 1; ii += 1
    b = inp[ii] - 1; ii += 1
    k = inp[ii]; ii += 1
 
    shortest_odd = 10**9 + 11
    shortest_even = 10**9 + 10
    for d in [dist(a,b), dist(a,x) + dist(y,b) + 1, dist(a,y) + dist(x,b) + 1]:
        
        if d & 1:
            shortest_odd = min(shortest_odd, d)
        else:
            shortest_even = min(shortest_even, d)
 
    if k & 1:
        out.append(b'YES' if shortest_odd <= k else b'NO')
    else:
        out.append(b'YES' if shortest_even <= k else b'NO')
os.write(1, b'\n'.join(out))
```