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;
long long int n, u, v, q, x, y, a, b, k, p[100007][22], d[100007];
vector<long long int> adj[100007];
void dfs(long long int x, long long int par) {
  d[x] = d[par] + 1;
  p[x][0] = par;
  for (long long int i = 1; i <= 20; ++i) {
    p[x][i] = p[p[x][i - 1]][i - 1];
  }
  for (long long int i : adj[x]) {
    if (i != par) dfs(i, x);
  }
}
long long int dist(long long int a, long long int b) {
  if (d[a] > d[b]) swap(a, b);
  long long int res = 0;
  for (long long int i = 20; i >= 0; --i) {
    if (d[p[b][i]] >= d[a]) {
      b = p[b][i];
      res += (1 << i);
    }
  }
  if (a == b) return res;
  for (long long int i = 20; i >= 0; --i) {
    if (p[b][i] != p[a][i]) {
      b = p[b][i];
      a = p[a][i];
      res += (1 << (i + 1));
    }
  }
  return res + 2;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long int i = 0; i < n - 1; ++i) {
    long long int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(1, 0);
  cin >> q;
  while (q--) {
    cin >> x >> y >> a >> b >> k;
    long long int ans = dist(a, b), ans2 = (dist(a, x) + dist(y, b)) + 1,
                  ans3 = (dist(a, y) + dist(x, b)) + 1;
    long long int f = 0x3f3f3f3f3f3f3f3f;
    if (ans % 2 == k % 2) f = min(f, ans);
    if (ans2 % 2 == k % 2) f = min(f, ans2);
    if (ans3 % 2 == k % 2) f = min(f, ans3);
    if (f <= k)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
