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 n, tim = 0;
long long strtim[100005], endtim[100005], level[100005];
long long par[20 + 5][100005];
vector<long long> gr[100005];
void dfs(long long x, long long parent, long long lvl) {
  strtim[x] = ++tim;
  level[x] = lvl;
  par[0][x] = parent;
  for (auto it : gr[x])
    if (it != parent) dfs(it, x, lvl + 1);
  endtim[x] = tim;
}
void precompute() {
  for (long long j = 1; j < 20; j++)
    for (long long i = 1; i <= n; i++) par[j][i] = par[j - 1][par[j - 1][i]];
}
long long kth_ancestor(long long u, long long k) {
  for (long long j = 20; j >= 0; j--)
    if (k & (1LL << j)) u = par[j][u];
  return u;
}
long long LCA(long long u, long long v) {
  if (level[u] < level[v]) swap(u, v);
  long long diff = level[u] - level[v];
  for (long long i = 20 - 1; i >= 0; i--) {
    if ((1 << i) & diff) {
      u = par[i][u];
    }
  }
  if (u == v) return u;
  for (long long i = 20 - 1; i >= 0; i--) {
    if (par[i][u] && par[i][u] != par[i][v]) {
      u = par[i][u];
      v = par[i][v];
    }
  }
  return par[0][u];
}
long long dist(long long u, long long v) {
  return level[u] + level[v] - 2 * level[LCA(u, v)];
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (long long i = 1; i < n; i++) {
    long long x, y;
    cin >> x >> y;
    gr[x].push_back(y);
    gr[y].push_back(x);
  }
  dfs(1, 0, 1);
  precompute();
  long long q;
  cin >> q;
  while (q--) {
    long long x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    long long flag = 0;
    long long dist_a_b = dist(a, b);
    long long dist_a_x_y_b = dist(a, x) + 1 + dist(y, b);
    long long dist_a_y_x_b = dist(a, y) + 1 + dist(x, b);
    if ((k - dist_a_b) >= 0 and (k - dist_a_b) % 2 == 0) flag = 1;
    if ((k - dist_a_x_y_b) >= 0 and (k - dist_a_x_y_b) % 2 == 0) flag = 1;
    if ((k - dist_a_y_x_b) >= 0 and (k - dist_a_y_x_b) % 2 == 0) flag = 1;
    if (flag == 1)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
