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;
inline void normal(int64_t &a) {
  a %= 1000000007;
  (a < 0) && (a += 1000000007);
}
inline int64_t modMul(int64_t a, int64_t b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline int64_t modAdd(int64_t a, int64_t b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline int64_t modSub(int64_t a, int64_t b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline int64_t modPow(int64_t b, int64_t p) {
  int64_t r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline int64_t modInverse(int64_t a) { return modPow(a, 1000000007 - 2); }
inline int64_t modDiv(int64_t a, int64_t b) { return modMul(a, modInverse(b)); }
template <typename F, typename S>
ostream &operator<<(ostream &os, const pair<F, S> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  typename vector<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  typename set<T>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename F, typename S>
ostream &operator<<(ostream &os, const map<F, S> &v) {
  os << "[";
  typename map<F, S>::const_iterator it;
  for (it = v.begin(); it != v.end(); it++) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
const int64_t siz = 2e5 + 1;
vector<int64_t> adj[siz];
int64_t timer, n, l;
vector<int64_t> in, out, level;
vector<vector<int64_t> > up;
void dfs(int64_t node, int64_t parent, int64_t h = 0) {
  in[node] = ++timer;
  up[node][0] = parent;
  level[node] = h;
  for (int64_t i = 1; i <= l; i++) {
    up[node][i] = up[up[node][i - 1]][i - 1];
  }
  for (auto to : adj[node]) {
    if (to != parent) {
      dfs(to, node, h + 1);
    }
  }
  out[node] = ++timer;
}
bool is_ancestor(int64_t u, int64_t v) {
  return (in[u] <= in[v] and out[u] >= out[v]);
}
int64_t lca(int64_t u, int64_t v) {
  if (is_ancestor(u, v)) return u;
  if (is_ancestor(v, u)) return v;
  for (int64_t i = l; i >= 0; i--) {
    if (!is_ancestor(up[u][i], v)) u = up[u][i];
  }
  return up[u][0];
}
void preprocess(int64_t root) {
  in.resize(n + 1);
  out.resize(n + 1);
  level.resize(n + 1);
  timer = 0;
  l = ceil(log2(n + 1));
  up.assign(n + 1, vector<int64_t>(l + 1));
  dfs(root, root);
}
int64_t path_length(int64_t u, int64_t v) {
  int64_t l = lca(u, v);
  return level[u] + level[v] - 2 * level[l];
}
bool check(int64_t x, int64_t k) {
  if (x <= k && x % 2 == k % 2) return true;
  return false;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int64_t T = 1;
  for (int64_t trr = 1; trr < T + 1; trr++) {
    int64_t q;
    cin >> n;
    for (int64_t i = 0; i < n - 1; i++) {
      int64_t u, v;
      cin >> u >> v;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    preprocess(1);
    bool wrong;
    cin >> q;
    while (q--) {
      int64_t x, y, a, b, k;
      cin >> x >> y >> a >> b >> k;
      int64_t ab = path_length(a, b), ax = path_length(a, x),
              yb = path_length(b, y), ay = path_length(a, y),
              bx = path_length(b, x);
      if (check(ab, k) || check(ax + yb + 1, k) || check(ay + bx + 1, k))
        cout << "YES" << endl;
      else
        cout << "NO" << endl;
    }
  }
}
