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;
using ll = long long;
using ld = long double;
constexpr ll inf = 0x3FFFFFFFFFFFFF;
struct SparseTable {
  vector<vector<int>> st;
  vector<int> log;
  vector<ll> *a;
  bool better(int lidx, int ridx) { return a->at(lidx) <= a->at(ridx); }
  void init(vector<ll> *vec) {
    a = vec;
    log.assign((ll)(*a).size() + 1, 0);
    for (int i = 2; i <= (ll)(*a).size(); i++) log[i] = log[i / 2] + 1;
    st.assign((ll)(*a).size(), vector<int>(log.back() + 1));
    for (int i = 0; i < (ll)(*a).size(); i++) st[i][0] = i;
    for (int j = 1; j <= log.back(); j++) {
      for (int i = 0; i + (1 << j) <= (ll)(*a).size(); i++) {
        st[i][j] = better(st[i][j - 1], st[i + (1 << (j - 1))][j - 1])
                       ? st[i][j - 1]
                       : st[i + (1 << (j - 1))][j - 1];
      }
    }
  }
  int queryIdempotent(int l, int r) {
    int j = log[r - l];
    return better(st[l][j], st[r - (1 << j)][j]) ? st[l][j]
                                                 : st[r - (1 << j)][j];
  }
};
struct LCA {
  vector<ll> depth;
  vector<int> visited, first;
  int idx;
  SparseTable st;
  void init(vector<vector<int>> &g, int root) {
    depth.assign(2 * (ll)(g).size(), 0);
    visited.assign(2 * (ll)(g).size(), -1);
    first.assign((ll)(g).size(), 2 * (ll)(g).size());
    idx = 0;
    visit(g, root);
    st.init(&depth);
  }
  void visit(vector<vector<int>> &g, int v, ll d = 0, int p = -1) {
    visited[idx] = v, depth[idx] = d;
    first[v] = min(idx, first[v]), idx++;
    for (int w : g[v]) {
      if (w != p && first[w] == 2 * (ll)(g).size()) {
        visit(g, w, d + 1, v);
        visited[idx] = v, depth[idx] = d, idx++;
      }
    }
  }
  int getLCA(int a, int b) {
    if (first[a] > first[b]) swap(a, b);
    return visited[st.queryIdempotent(first[a], first[b] + 1)];
  }
  int getDist(int a, int b) {
    int lca = getLCA(a, b);
    return abs(depth[first[a]] - depth[first[lca]]) +
           abs(depth[first[b]] - depth[first[lca]]);
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  vector<vector<int>> adj(n);
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  LCA lca;
  lca.init(adj, 0);
  int q;
  cin >> q;
  for (int i = 0; i < q; i++) {
    ll x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    x--;
    y--;
    a--;
    b--;
    bool ok = false;
    auto check = [&](int d) {
      if (d <= k && (k - d) % 2 == 0) ok = true;
    };
    check(lca.getDist(a, b));
    check(lca.getDist(a, x) + 1 + lca.getDist(y, b));
    check(lca.getDist(a, y) + 1 + lca.getDist(x, b));
    cout << (ok ? "YES" : "NO") << endl;
  }
}
