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>
const long long INF = 1e18 + 7, mod = 998244353, MAXN = 1e5, SQ = 450,
                pp = 1e5 + 3;
const long double EPS = 1e-9, PI = acos(-1);
using namespace std;
namespace in_out {
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2> a) {
  out << a.first << " " << a.second;
  return out;
}
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& a) {
  in >> a.first >> a.second;
  return in;
}
template <typename T>
ostream& operator<<(ostream& out, vector<T> a) {
  for (auto i : a) out << i << " ";
  return out;
}
template <typename T>
istream& operator>>(istream& in, vector<T>& a) {
  for (auto& i : a) in >> i;
  return in;
}
}  // namespace in_out
using namespace in_out;
vector<vector<int>> g, up;
vector<int> deep;
int clca, clen;
int x, y, a, b, k;
void dfs(int v, int p = -1, int d = 0) {
  if (p != -1) {
    up[0][v] = p;
    for (int i = 1; i < 25; i++) up[i][v] = up[i - 1][up[i - 1][v]];
  }
  deep[v] = d;
  for (auto& i : g[v])
    if (i != p) dfs(i, v, d + 1);
}
int get_lca(int v, int u) {
  if (deep[v] < deep[u]) swap(v, u);
  int h = deep[v] - deep[u];
  for (int i = 24; i >= 0; i--)
    if (h >= (1 << i)) v = up[i][v], h -= (1 << i);
  for (int i = 24; i >= 0; i--) {
    if (up[i][v] == up[i][u]) continue;
    v = up[i][v];
    u = up[i][u];
  }
  if (v != u) return up[0][v];
  return v;
}
int dist(int v, int u, int lca) { return deep[v] + deep[u] - 2 * deep[lca]; }
signed main() {
  ios_base::sync_with_stdio();
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  g.resize(n);
  deep.resize(n);
  up.resize(25, vector<int>(n));
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    cin >> v >> u;
    v--;
    u--;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  dfs(0);
  int q;
  cin >> q;
  while (q--) {
    cin >> x >> y >> a >> b >> k;
    x--;
    y--;
    a--;
    b--;
    int d1 = dist(a, b, get_lca(a, b));
    int d2 = dist(a, x, get_lca(a, x)) + dist(b, y, get_lca(b, y)) + 1;
    int d3 = dist(b, x, get_lca(b, x)) + dist(a, y, get_lca(a, y)) + 1;
    if ((d1 <= k && d1 % 2 == k % 2) || (d2 <= k && d2 % 2 == k % 2) ||
        (d3 <= k && d3 % 2 == k % 2))
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}
