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;
class LCA {
 private:
  vector<vector<int>> anc;
  vector<vector<int>> cost;
  vector<int> level;
  vector<int> dist;
  vector<bool> vis;
  int start_idx;
  int n;

 private:
  void build(const vector<vector<pair<int, int>>> &adj) {
    dist[this->start_idx] = 0;
    dfs_LCA(this->start_idx, -1, 1, -1, adj);
    for (int j = 1; j < anc.front().size(); j++)
      for (int i = 0; i < anc.size(); i++)
        if (this->anc[i][j - 1] != -1) {
          this->anc[i][j] = this->anc[this->anc[i][j - 1]][j - 1];
        }
  }
  void dfs_LCA(const int u, const int p, const int d, const int w,
               const vector<vector<pair<int, int>>> &adj) {
    this->level[u] = d;
    this->vis[u] = true;
    this->anc[u][0] = p;
    for (const pair<int, int> &x : adj[u]) {
      if (this->vis[x.first]) continue;
      this->dist[x.first] = this->dist[u] + x.second;
      dfs_LCA(x.first, u, d + 1, x.second, adj);
    }
  }

 public:
  explicit LCA(const int n, const int start_idx,
               const vector<vector<pair<int, int>>> &adj) {
    this->n = n;
    this->anc.resize(n + 1, vector<int>((int)log2(n) + 1, -1));
    this->level.resize(n + 1, -1);
    this->dist.resize(n + 1);
    this->vis.resize(n + 1, false);
    this->start_idx = start_idx;
    this->build(adj);
  }
  int query(int a, int b) {
    assert(start_idx <= min(a, b));
    assert(max(a, b) < this->n + start_idx);
    const int a_cpy = a, b_cpy = b;
    if (this->level[b] > this->level[a]) swap(a, b);
    const int lg = log2(this->level[a]);
    for (int i = lg; i >= 0; i--)
      if (this->level[a] - (1 << i) >= this->level[b]) a = this->anc[a][i];
    if (a == b)
      return this->dist[a_cpy] + this->dist[b_cpy] - 2 * this->dist[a];
    for (int i = lg; i >= 0; i--)
      if (this->anc[a][i] != -1 && this->anc[a][i] != this->anc[b][i]) {
        a = this->anc[a][i];
        b = this->anc[b][i];
      }
    return this->dist[a_cpy] + this->dist[b_cpy] - 2 * this->dist[anc[a][0]];
  }
  int max_edge(int v, int p) {
    if (this->level[v] < this->level[p]) swap(v, p);
    int k = this->level[v] - this->level[p];
    int ans = -1;
    for (int i = 0; k > 0; i++, k >>= 1)
      if (k & 1) {
        v = this->anc[v][i];
      }
    return ans;
  }
  int go_back(int x, int k) {
    for (int i = 0; k > 0; i++, k >>= 1)
      if (k & 1) x = this->anc[x][i];
    return x;
  }
};
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<vector<pair<int, int>>> adj(n + 1);
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(pair<int, int>(v, 1));
    adj[v].push_back(pair<int, int>(u, 1));
  }
  LCA lca(n, 1, adj);
  int q;
  cin >> q;
  while (q--) {
    int from, to, x, y, k;
    cin >> x >> y >> from >> to >> k;
    string ans = "NO";
    int p1 = lca.query(from, to);
    if (p1 <= k && ((p1 & 1) == (k & 1))) ans = "YES";
    int p2 = lca.query(from, y) + 1 + lca.query(x, to);
    if (p2 <= k && ((p2 & 1) == (k & 1))) ans = "YES";
    int p3 = lca.query(from, x) + 1 + lca.query(y, to);
    if (p3 <= k && ((p3 & 1) == (k & 1))) ans = "YES";
    cout << ans << '\n';
  }
}
