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;
int first[(100 * 1000 + 5)], height[(100 * 1000 + 5)], tim = 0;
vector<int> euler, g[(100 * 1000 + 5)];
void dfs(int u, int p = -1, int h = 0) {
  first[u] = tim++;
  euler.push_back(u);
  height[u] = h;
  for (int v : g[u]) {
    if (v != p) {
      dfs(v, u, h + 1);
      euler.push_back(u);
      tim++;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1);
  int sz = 1;
  int m = euler.size();
  while (sz <= m) {
    sz *= 2;
  }
  vector<pair<int, int>> tree(2 * sz, {((int)1e9), -1});
  for (int i = 0; i < m; i++) {
    tree[i + sz] = make_pair(height[euler[i]], euler[i]);
  }
  for (int i = sz - 1; i > 0; i--) {
    tree[i] = min(tree[2 * i], tree[2 * i + 1]);
  }
  auto lca = [&](int u, int v) {
    int L = first[u] + sz, R = first[v] + sz;
    assert(L && R);
    if (L > R) swap(L, R);
    pair<int, int> ans = {((int)1e9), -1};
    while (L <= R) {
      if (L % 2 == 1) ans = min(ans, tree[L++]);
      if (R % 2 == 0) ans = min(ans, tree[R--]);
      L /= 2, R /= 2;
    }
    assert(ans.second != -1);
    return ans.second;
  };
  auto dist = [&](int u, int v) {
    return height[u] + height[v] - 2 * height[lca(u, v)];
  };
  int queries;
  cin >> queries;
  while (queries--) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    vector<int> options = {dist(a, b), dist(a, x) + dist(b, y) + 1,
                           dist(a, y) + dist(b, x) + 1};
    bool ok = false;
    for (int d : options) {
      ok |= d <= k && d % 2 == k % 2;
    }
    cout << (ok ? "YES" : "NO") << '\n';
  }
}
