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;
long long int M = 1e9 + 7;
long long int inf = 9 * 1e18;
long long int begtime = clock();
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
vector<string> vec_splitter(string s) {
  s += ',';
  vector<string> res;
  while (!s.empty()) {
    res.push_back(s.substr(0, s.find(',')));
    s = s.substr(s.find(',') + 1);
  }
  return res;
}
void debug_out(vector<string> __attribute__((unused)) args,
               __attribute__((unused)) int idx,
               __attribute__((unused)) int LINE_NUM) {
  cout << endl;
}
template <typename Head, typename... Tail>
void debug_out(vector<string> args, int idx, int LINE_NUM, Head H, Tail... T) {
  if (idx > 0)
    cout << " | ";
  else
    cout << "Line(" << LINE_NUM << ") ";
  stringstream ss;
  ss << H;
  cout << args[idx] << " = " << ss.str();
  debug_out(args, idx + 1, LINE_NUM, T...);
}
long long int n, m;
long long int binpow(long long int val, long long int deg) {
  if (deg < 0) return 0;
  if (!deg) return 1 % M;
  if (deg & 1) return binpow(val, deg - 1) * val % M;
  long long int res = binpow(val, deg >> 1);
  return (res * res) % M;
}
long long int modinv(long long int n) { return binpow(n, M - 2); }
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long int l;
vector<vector<long long int>> adj;
long long int timer;
vector<long long int> tin, tout;
vector<vector<long long int>> up;
void dfs(long long int vector, long long int p) {
  tin[vector] = ++timer;
  up[vector][0] = p;
  for (long long int i = 1; i <= l; ++i)
    up[vector][i] = up[up[vector][i - 1]][i - 1];
  for (long long int u : adj[vector]) {
    if (u != p) dfs(u, vector);
  }
  tout[vector] = ++timer;
}
bool is_ancestor(long long int u, long long int vector) {
  return tin[u] <= tin[vector] && tout[u] >= tout[vector];
}
long long int lca(long long int u, long long int vector) {
  if (is_ancestor(u, vector)) return u;
  if (is_ancestor(vector, u)) return vector;
  for (long long int i = l; i >= 0; --i) {
    if (!is_ancestor(up[u][i], vector)) u = up[u][i];
  }
  return up[u][0];
}
void preprocess(long long int root) {
  tin.resize(n);
  tout.resize(n);
  timer = 0;
  l = ceil(log2(n));
  up.assign(n, vector<long long int>(l + 1));
  dfs(root, root);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int i, j, t, k, x, y, z, N;
  cin >> n;
  adj.resize(n);
  for (i = 0; i < n - 1; i++) {
    cin >> x >> y;
    x--;
    y--;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  preprocess(0);
  queue<long long int> q;
  vector<long long int> used(n, 0);
  vector<long long int> d(n, inf), p(n, -1);
  long long int s = 0;
  q.push(s);
  used[s] = 1;
  d[s] = 0;
  while (!q.empty()) {
    long long int V = q.front();
    q.pop();
    for (long long int u : adj[V]) {
      if (!used[u]) {
        used[u] = 1;
        q.push(u);
        d[u] = d[V] + 1;
        p[u] = V;
      }
    }
  }
  long long int m;
  cin >> m;
  for (i = 0; i < m; i++) {
    long long int a, b;
    cin >> x >> y >> a >> b >> k;
    x--;
    y--;
    a--;
    b--;
    long long int d1 = d[a] + d[b] - 2 * d[lca(a, b)];
    long long int flg = 0;
    if (d1 <= k) {
      if ((k - d1) % 2 == 0) flg = 1;
    }
    long long int d2 = d[x] + d[a] - 2 * d[lca(a, x)];
    long long int d3 = d[y] + d[b] - 2 * d[lca(b, y)];
    long long int dum = d2 + d3 + 1;
    if (dum <= k) {
      if ((k - dum) % 2 == 0) flg = 1;
    }
    d2 = d[x] + d[b] - 2 * d[lca(b, x)];
    d3 = d[y] + d[a] - 2 * d[lca(a, y)];
    dum = d2 + d3 + 1;
    if (dum <= k) {
      if ((k - dum) % 2 == 0) flg = 1;
    }
    if (flg)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
