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 
n = int(input())
g = [[] for _ in range(n)]
for i in range(n - 1):
    u, v = map(int, input().split())
    g[u - 1].append(v - 1)
    g[v - 1].append(u - 1)

logn = 19
st = [0] * n
dep = [-1] * n
par = [[0] * n for _ in range(logn + 1)]

st[0], cur = 0, 1
dep[0] = 0
while cur > 0: 
    x = st[cur - 1]
    cur = cur - 1
    for i in g[x]:
        if dep[i] is -1:
            dep[i] = 1 + dep[x]
            par[0][i] = x
            st[cur] = i
            cur = cur + 1

for i in range(1, logn + 1):
    for j in range(n):
        par[i][j] = par[i - 1][par[i - 1][j]]

def lca(i, j):
    if dep[i] > dep[j]:
        return lca(j, i)
    var = logn
    while var >= 0: 
        if dep[j] - (1 << var) >= dep[i]:
            j = par[var][j]
        var = var - 1
    if i == j: return i
    var = logn
    while var >= 0:
        if par[var][i] is not par[var][j]:
            i = par[var][i]
            j = par[var][j]
        var = var - 1
    return par[0][i]

def dist(i, j):
    return dep[i] + dep[j] - 2 * dep[lca(i, j)]


def query(a, b, x, y, k):
    can = []
    can.append(dist(x, y))
    can.append(dist(x, a) + 1 + dist(b, y))
    can.append(dist(x, b) + 1 + dist(a, y))
    for i in can: 
        if i % 2 == k % 2 and i <= k:
            return 'YES'
    return 'NO'

q = int(input())
res = []
for i in range(q):
    x, y, a, b, k = map(int, input().split())
    res.append(query(x - 1, y - 1, a - 1, b - 1, k))
print('\n'.join(res))
