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 mod1 = 998244353, mod = 1e9 + 7;
const int MAXN = 1e6 + 6, MAXM = 2e5 + 5;
const int inf = 2e9;
const long long linf = 1e18;
template <class T>
struct RMQ {
  vector<vector<T>> spt;
  vector<int> lg;
  void init(const vector<T> &arr) {
    int N = arr.size(), tmp = 1, K = 1;
    while (tmp < N) tmp *= 2, ++K;
    lg.assign(N, 0);
    spt.assign(K, arr);
    for (int i = 2; i < N; i++) lg[i] = lg[i / 2] + 1;
    for (int i = 1; i < K; i++) {
      for (int j = 0; j < N; j++) {
        spt[i][j] =
            min(spt[i - 1][j], spt[i - 1][min(N - 1, j + (1 << (i - 1)))]);
      }
    }
  }
  T query(int l, int r) {
    int d = lg[r - l];
    return min(spt[d][l], spt[d][r - (1 << d)]);
  }
};
struct LCA {
  vector<int> st;
  vector<int> dep;
  vector<pair<int, int>> arr;
  int N;
  RMQ<pair<int, int>> rmq;
  LCA(vector<vector<int>> &adj, int r = 0) {
    N = adj.size();
    st.resize(N);
    dep.resize(N);
    arr.reserve(N * 2);
    dfs(adj, r, 0);
    rmq.init(arr);
  }
  void dfs(vector<vector<int>> &adj, int u, int p, int h = 0) {
    st[u] = arr.size();
    dep[u] = h;
    arr.emplace_back(h, u);
    for (auto v : adj[u])
      if (v != p) {
        dfs(adj, v, u, h + 1);
        arr.emplace_back(h, u);
      }
  }
  int query(int a, int b) {
    if (a == b) return a;
    a = st[a], b = st[b];
    return rmq.query(min(a, b), max(a, b)).second;
  }
  int dist(int a, int b) {
    int lca = query(a, b);
    return dep[a] + dep[b] - 2 * dep[lca];
  }
};
void Solve() {
  int n;
  cin >> n;
  vector<vector<int>> adj(n, vector<int>());
  for (int i = 1, u, v; i < n; i++) {
    cin >> u >> v;
    u--;
    v--;
    adj[u].emplace_back(v);
    adj[v].emplace_back(u);
  }
  LCA lca(adj);
  int q, x, y, a, b, k;
  cin >> q;
  while (q--) {
    cin >> x >> y >> a >> b >> k;
    x--;
    y--;
    a--;
    b--;
    int d1 = lca.dist(a, b);
    int d2 = lca.dist(a, x) + lca.dist(b, y) + 1;
    int d3 = lca.dist(a, y) + lca.dist(b, x) + 1;
    if ((d1 <= k && (k - d1) % 2 == 0) || (d2 <= k && (k - d2) % 2 == 0) ||
        (d3 <= k && (k - d3) % 2 == 0)) {
      cout << "YES\n";
    } else {
      cout << "NO\n";
    }
  }
}
signed main() {
  if (fopen(""
            ".inp",
            "r")) {
    freopen(
        ""
        ".inp",
        "r", stdin);
    freopen(
        ""
        ".out",
        "w", stdout);
  };
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int TC = 1;
  while (TC--) Solve();
  return 0;
}
