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>
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")
const int N = 2e5 + 5;
const long long int mod = 1e9 + 7;
const long long int Mod = 998244353;
const long double Pi = acos(-1);
const long long int Inf = 4e18;
using namespace std;
struct LCA {
  vector<int> euler, d, f, t, g[100001];
  vector<bool> vis;
  int n;
  void Initialize(int s) {
    n = s;
    d.assign(n + 1, INT_MAX);
    f.assign(n + 1, INT_MAX);
    vis.assign(n + 1, false);
    d[1] = 0;
    t.assign(3 * N, 0);
  }
  void Dfs(int t) {
    vis[t] = true;
    euler.push_back(t);
    for (auto u : g[t]) {
      if (!vis[u]) {
        d[u] = d[t] + 1;
        Dfs(u);
        euler.push_back(t);
      }
    }
  }
  void Update(int idx, int node, int v, int tl, int tr) {
    if (tl == tr)
      t[v] = node;
    else {
      int tm = (tl + tr) / 2;
      if (idx <= tm)
        Update(idx, node, 2 * v + 1, tl, tm);
      else
        Update(idx, node, 2 * v + 2, tm + 1, tr);
      t[v] = (d[t[2 * v + 1]] > d[t[2 * v + 2]] ? t[2 * v + 2] : t[2 * v + 1]);
    }
  }
  int Query(int l, int r, int v, int tl, int tr) {
    if (tl == l && tr == r)
      return t[v];
    else {
      int tm = (tl + tr) / 2;
      if (r <= tm)
        return Query(l, r, 2 * v + 1, tl, tm);
      else if (l > tm)
        return Query(l, r, 2 * v + 2, tm + 1, tr);
      else {
        int L = Query(l, tm, 2 * v + 1, tl, tm);
        int R = Query(tm + 1, r, 2 * v + 2, tm + 1, tr);
        if (d[L] < d[R])
          return L;
        else
          return R;
      }
    }
  }
  int AncestOR(int x, int y, int k) {
    x = f[x], y = f[y];
    if (x <= y)
      return Query(x, y, 0, 0, k - 1);
    else
      return Query(y, x, 0, 0, k - 1);
  }
} lca;
int Dist(int x, int y, int k) {
  return lca.d[x] + lca.d[y] - 2 * lca.d[lca.AncestOR(x, y, k)];
}
bool Check(int x, int y) { return (x <= y && (x % 2 == y % 2)); }
void TestCase() {
  int n, q;
  cin >> n;
  lca.Initialize(n);
  for (int i = 1, u, v; i < n; i++) {
    cin >> u >> v;
    lca.g[u].push_back(v);
    lca.g[v].push_back(u);
  }
  lca.Dfs(1);
  for (int i = 0; i < (int)lca.euler.size(); i++)
    lca.f[lca.euler[i]] = min(lca.f[lca.euler[i]], i);
  int m = (int)lca.euler.size(), msb = -1;
  for (int i = 31; i >= 0; i--) {
    if (m & (1 << i)) msb = max(msb, i);
  }
  m = (1 << (msb + 1));
  while ((int)lca.euler.size() < m) lca.euler.push_back(0);
  for (int i = 0; i < (int)lca.euler.size(); i++)
    lca.Update(i, lca.euler[i], 0, 0, m - 1);
  cin >> q;
  while (q--) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    bool ok = false;
    int ab = Dist(a, b, m), ax = Dist(a, x, m), ay = Dist(a, y, m);
    int xy = Dist(x, y, m), bx = Dist(b, x, m), by = Dist(b, y, m);
    ok |= Check(ab, k);
    ok |= Check(ax + xy + by, k);
    ok |= Check(ax + 1 + by, k);
    ok |= Check(ay + xy + bx, k);
    ok |= Check(ay + 1 + bx, k);
    ok |= Check(ax + (xy + 1) + xy + by, k);
    ok |= Check(ax + (xy + 1) + 1 + by, k);
    ok |= Check(ay + (xy + 1) + xy + bx, k);
    ok |= Check(ay + (xy + 1) + 1 + bx, k);
    ok ? cout << "YES\n" : cout << "NO\n";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  int T = 1;
  while (T--) {
    TestCase();
    cout << "\n";
  }
  return 0;
}
