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 
#include <bits/stdc++.h>
using std::cin;
using std::cout;
using std::string;
int n, m, d;
int ver[200010], next[200010], head[200010], tot;
int dis[100010], fa[100010][20];
inline void link(int x, int y) {
  ver[++tot] = y, next[tot] = head[x], head[x] = tot;
}
inline void dfs(int now, int father) {
  dis[now] = dis[father] + 1, fa[now][0] = father;
  for (register int i = head[now]; i; i = next[i]) {
    if (ver[i] != father) dfs(ver[i], now);
  }
  return;
}
inline int LCA(int x, int y) {
  if (dis[x] < dis[y]) std::swap(x, y);
  int tx = x, ty = y;
  for (register int i = d; i >= 0; --i) {
    if (dis[fa[x][i]] >= dis[y]) x = fa[x][i];
  }
  if (x != y) {
    for (register int i = d; i >= 0; --i) {
      if (fa[x][i] != fa[y][i]) x = fa[x][i], y = fa[y][i];
    }
    x = fa[x][0];
  }
  return std::abs(dis[x] - dis[tx]) + std::abs(dis[x] - dis[ty]);
}
int main() {
  std::ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int x, y, a, b, l;
  cin >> n;
  for (register int i = 1; i < n; ++i) {
    cin >> x >> y;
    link(x, y), link(y, x);
  }
  while ((1 << d) <= n) ++d;
  dfs(x, 0);
  for (register int j = 1; j <= d; ++j) {
    for (register int i = 1; i <= n; ++i) {
      fa[i][j] = fa[fa[i][j - 1]][j - 1];
    }
  }
  cin >> m;
  for (register int i = 1; i <= m; ++i) {
    cin >> x >> y >> a >> b >> l;
    int l1 = LCA(a, b), l2 = LCA(a, x) + LCA(b, y) + 1,
        l3 = LCA(b, x) + LCA(a, y) + 1;
    if (l1 <= l && l1 % 2 == l % 2)
      cout << "YES\n";
    else if (l2 <= l && l2 % 2 == l % 2)
      cout << "YES\n";
    else if (l3 <= l && l3 % 2 == l % 2)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
