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;
template <class T>
ostream& operator<<(ostream& os, vector<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << "]";
}
template <class L, class R>
ostream& operator<<(ostream& os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
const long long N = 1e6 + 5;
const long long logN = 22;
vector<long long> adj[N];
long long par[N][logN], lvl[N];
void dfs(long long f, long long p) {
  lvl[f] = lvl[p] + 1;
  par[f][0] = p;
  for (long long i = 1; i < logN; i++) {
    par[f][i] = par[par[f][i - 1]][i - 1];
  }
  for (auto z : adj[f])
    if (z - p) dfs(z, f);
}
long long lca(long long u, long long v) {
  if (lvl[v] < lvl[u]) swap(u, v);
  for (long long i = logN - 1; i >= 0; i--) {
    if (lvl[par[v][i]] >= lvl[u]) v = par[v][i];
  }
  if (v == u) return v;
  for (long long i = logN - 1; i >= 0; i--) {
    if (par[v][i] != par[u][i]) v = par[v][i], u = par[u][i];
  }
  return par[v][0];
}
long long dist(long long u, long long v) {
  long long l = lca(u, v);
  return -2 * lvl[l] + lvl[u] + lvl[v];
}
long long gcd(long long a, long long b, long long& x, long long& y) {
  if (a == 0) {
    x = 0;
    y = 1;
    return b;
  }
  long long x1, y1;
  long long d = gcd(b % a, a, x1, y1);
  x = y1 - (b / a) * x1;
  y = x1;
  return d;
}
bool find_any_solution(long long a, long long b, long long c, long long& x0,
                       long long& y0, long long& g) {
  g = gcd(abs(a), abs(b), x0, y0);
  if (c % g) {
    return false;
  }
  x0 *= c / g;
  y0 *= c / g;
  if (a < 0) x0 = -x0;
  if (b < 0) y0 = -y0;
  return true;
}
const long long inf = 1e13;
bool poss(long long al, long long c, long long k) {
  long long x0, y0, g;
  bool go = find_any_solution(2, c, k - al, x0, y0, g);
  if (!go) return 0;
  long long a, b;
  a = 2;
  b = c;
  assert(x0 * a + y0 * b == k - al);
  a = 2 / g;
  b = c / g;
  if (x0 >= 0 && y0 >= 0) return 1;
  if (x0 < 0 && y0 < 0) return 0;
  if (x0 < 0) {
    long long i = (-x0 + b - 1) / b;
    x0 += i * b;
    y0 -= i * a;
    if (x0 >= 0 && y0 >= 0) return 1;
    return 0;
  }
  swap(x0, y0);
  swap(a, b);
  if (x0 < 0) {
    long long i = (-x0 + b - 1) / b;
    x0 += i * b;
    y0 -= i * a;
    if (x0 >= 0 && y0 >= 0) return 1;
    return 0;
  }
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(25);
  long long n, q, a, b, c;
  cin >> n;
  for (long long i = 1; i < n; i++) {
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  lvl[1] = -1;
  dfs(1, 1);
  cin >> q;
  while (q--) {
    long long x, y, k;
    cin >> x >> y >> a >> b >> k;
    long long al1 = dist(a, b);
    long long al2 = dist(a, x) + 1 + dist(y, b);
    long long al3 = dist(a, y) + 1 + dist(x, b);
    al3 = min(al3, al2);
    if (al1 <= k && al1 % 2 == k % 2)
      cout << "YES\n";
    else if (al3 <= k && al3 % 2 == k % 2)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
}
