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 
def lca(u, v):
    if h[u] < h[v]:
        u, v = v, u
    for i in range(h[u].bit_length() - 1, -1, -1):
        if par[u][i] != -1 and h[par[u][i]] >= h[v]:
            u = par[u][i]
    if u == v:
        return u
    for i in range(h[u].bit_length() - 1, -1, -1):
        if par[u][i] != par[v][i]:
            u = par[u][i]
            v = par[v][i]
    return par[u][0]


def dist(u, v):
    lc = lca(u, v)
    return h[u] - h[lc] + h[v] - h[lc]


def cal(x, y, a, b, k):
    d = dist(a, b)
    if d <= k and (k - d) & 1 == 0:
        return True
    d = dist(a, x) + 1 + dist(y, b)
    if d <= k and (k - d) & 1 == 0:
        return True
    d = dist(a, y) + 1 + dist(x, b)
    if d <= k and (k - d) & 1 == 0:
        return True


def dfs(u, p=-1):
    stack = [(u, p)]
    while stack:
        u, p = stack.pop()
        par[u][0] = p
        if p != -1:
            h[u] = h[p] + 1
        for i in range(1, LN):
            if par[u][i - 1] != -1:
                par[u][i] = par[par[u][i - 1]][i - 1]
        for v in adj[u]:
            if v != p:
                stack.append((v, u))


n = int(input())
h = [0] * n
LN = (n - 1).bit_length()
par = [[-1] * LN for _ in range(n)]
adj = [[] for _ in range(n)]
vs = (map(int, input().split()) for _ in range(n - 1))
for u, v in vs:
    u -= 1
    v -= 1
    adj[u].append(v)
    adj[v].append(u)
dfs(0)

q = int(input())
for _ in range(q):
    x, y, a, b, k = map(int, input().split())
    print('YES' if cal(x - 1, y - 1, a - 1, b - 1, k) else 'NO')
