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;
const int N = 1e5 + 5;
vector<int> adj[N];
int lvl[N], anc[N][25];
int n;
void dfs(int u = 0, int par = 0, int l = 0) {
  lvl[u] = l;
  anc[u][0] = par;
  for (auto nxt : adj[u])
    if (nxt != par) dfs(nxt, u, l + 1);
}
void buildLCA() {
  int lg = ceil(log2(n));
  for (int j = 1; j < lg; j++)
    for (int i = 0; i < n; i++) anc[i][j] = anc[anc[i][j - 1]][j - 1];
}
int LCA(int i, int j) {
  int lg = ceil(log2(n));
  int st = lg;
  if (lvl[i] > lvl[j]) swap(i, j);
  int cur = lvl[j];
  for (; st >= 0; st--)
    if (cur - (1 << st) >= lvl[i]) cur -= (1 << st), j = anc[j][st];
  if (i == j) return 2 * i - j;
  cur = lvl[i];
  for (st = lg; st >= 0; st--)
    if (anc[i][st] != anc[j][st])
      cur -= (1 << st), i = anc[i][st], j = anc[j][st];
  return anc[i][0];
}
int dis(int a, int b) { return lvl[a] + lvl[b] - 2 * lvl[LCA(a, b)]; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> n;
  int a, b;
  for (int i = 0; i < int(n - 1); i++) {
    cin >> a >> b;
    a--, b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs();
  buildLCA();
  cin >> q;
  while (q--) {
    int x, y, a, b, k;
    cin >> a >> b >> x >> y >> k;
    x--, y--, a--, b--;
    {
      int d = dis(x, y);
      if (d <= k && (k - d) % 2 == 0) {
        cout << "YES" << '\n';
        continue;
      }
    }
    {
      int d = dis(x, a) + dis(y, b) + 1;
      if (d <= k && (k - d) % 2 == 0) {
        cout << "YES" << '\n';
        continue;
      }
    }
    {
      int d = dis(x, b) + dis(y, a) + 1;
      if (d <= k && (k - d) % 2 == 0) {
        cout << "YES" << '\n';
        continue;
      }
    }
    cout << "NO" << '\n';
  }
  return 0;
}
