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 namespace std;
struct Edge {
  int nxt, to;
};
int n, m, s;
Edge e[100005 << 1];
int first[100005];
int ec;
int fa[100005][21];
int dep[100005];
int _log[100005];
void addEdge(int from, int to) {
  e[++ec].to = to;
  e[ec].nxt = first[from];
  first[from] = ec;
}
void dfs(int u, int f) {
  fa[u][0] = f;
  dep[u] = dep[f] + 1;
  for (int i = 1; i <= _log[dep[u]]; ++i) {
    fa[u][i] = fa[fa[u][i - 1]][i - 1];
  }
  for (int i = first[u]; i; i = e[i].nxt) {
    if (e[i].to != f) dfs(e[i].to, u);
  }
}
int lca(int a, int b) {
  if (dep[a] < dep[b]) swap(a, b);
  while (dep[a] > dep[b]) {
    a = fa[a][_log[dep[a] - dep[b]] - 1];
  }
  if (a == b) return a;
  for (int i = _log[dep[a]] - 1; i >= 0; --i) {
    if (fa[a][i] != fa[b][i]) a = fa[a][i], b = fa[b][i];
  }
  return fa[a][0];
}
inline int dis(int x, int y) { return dep[x] + dep[y] - dep[lca(x, y)] * 2; }
int solve(int x, int y, int a, int b, int k) {
  int tmp = dis(a, b);
  if ((tmp <= k) && (tmp - k) % 2 == 0) return 1;
  tmp = dis(a, x) + 1 + dis(y, b);
  if ((tmp <= k) && (tmp - k) % 2 == 0) return 1;
  tmp = dis(a, y) + 1 + dis(x, b);
  if ((tmp <= k) && (tmp - k) % 2 == 0) return 1;
  return 0;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    _log[i] = _log[i - 1] + ((1 << _log[i - 1]) == i);
  }
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    addEdge(u, v);
    addEdge(v, u);
  }
  dfs(1, 0);
  cin >> m;
  for (int i = 1; i <= m; ++i) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    cout << (solve(x, y, a, b, k) ? "YES" : "NO") << endl;
  }
  return 0;
}
