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>
const int inf = 1e9;
const long long INF = 1e15;
template <typename A0>
void print(A0 a0) {
  std::cout << a0 << '\n';
}
template <typename A0, typename... ARGS>
void print(A0 a0, ARGS... rest) {
  std::cout << a0 << ' ';
  print(rest...);
}
int dx[] = {0, 0, -1, 1}, dy[] = {1, -1, 0, 0};
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  vector<vector<int> > e(n);
  for (int i = 0; i < (n - 1); ++i) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    e[u].push_back(v);
    e[v].push_back(u);
  }
  vector<int> parent(n), depth(n);
  depth[0] = 0;
  function<void(int, int)> dfs = [&](int here, int p) {
    parent[here] = p;
    for (auto next : e[here])
      if (next != p) {
        depth[next] = depth[here] + 1;
        dfs(next, here);
      }
  };
  dfs(0, 0);
  vector<vector<int> > sparse;
  for (int k = 0; (1 << k) < n; k++) {
    if (k == 0) {
      sparse.emplace_back((parent).begin(), (parent).end());
      continue;
    }
    vector<int> temp(n);
    for (int i = 0; i < (n); ++i) temp[i] = sparse[k - 1][sparse[k - 1][i]];
    sparse.emplace_back((temp).begin(), (temp).end());
  }
  auto dist = [&](int u, int v) {
    if (depth[u] > depth[v]) swap(u, v);
    int up = depth[v] - depth[u];
    for (int k = 0; up >= (1 << k); k++) {
      if (up & (1 << k)) v = sparse[k][v];
    }
    if (u == v) {
      return up;
    }
    int up2 = 0;
    for (int k = sparse.size() - 1; k >= 0; k--) {
      if (sparse[k][u] != sparse[k][v]) {
        u = sparse[k][u];
        v = sparse[k][v];
        up2 += 2 * (1 << k);
      }
    }
    return up + up2 + 2;
  };
  int q;
  cin >> q;
  while (q--) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    x--, y--, a--, b--;
    int one = dist(a, b);
    int two = min(dist(a, x) + dist(y, b) + 1, dist(a, y) + 1 + dist(x, b));
    bool yes = 0;
    if (k >= one && (k - one) % 2 == 0) yes = 1;
    if (k >= two && (k - two) % 2 == 0) yes = 1;
    if (yes)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
}
