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>
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
constexpr int maxl = 21;
signed main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<vector<int>> graph(n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  vector<array<int, maxl>> step(n);
  vector<int> tin(n), tout(n), depth(n, -1);
  int time = 0;
  function<void(int, int)> dfs = [&](int v, int p) {
    step[v][0] = p;
    depth[v] = depth[p] + 1;
    tin[v] = time++;
    for (int ne : graph[v])
      if (ne != p) dfs(ne, v);
    tout[v] = time++;
  };
  dfs(0, 0);
  for (int l = 0; l < maxl - 1; l++)
    for (int v = 0; v < n; v++) step[v][l + 1] = step[step[v][l]][l];
  auto anc = [&](int high, int low) {
    return tin[high] <= tin[low] and tout[low] <= tout[high];
  };
  auto lca = [&](int u, int v) {
    if (anc(u, v)) return u;
    if (anc(v, u)) return v;
    for (int i = maxl - 1; i >= 0; i--)
      if (!anc(step[u][i], v)) u = step[u][i];
    return step[u][0];
  };
  auto dist = [&](int u, int v) {
    return depth[u] + depth[v] - 2 * depth[lca(u, v)];
  };
  int tt;
  cin >> tt;
  while (tt--) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    x--;
    y--;
    a--;
    b--;
    vector<int> ds = {dist(a, b), dist(a, x) + dist(y, b) + 1,
                      dist(b, x) + dist(y, a) + 1};
    bool gud = false;
    for (int d : ds) gud |= d % 2 == k % 2 and d <= k;
    cout << (gud ? "YES" : "NO") << "\n";
  }
}
