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 <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
template <typename T, typename U>
static inline void amin(T& x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T& x, U y) {
  if (x < y) x = y;
}
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long min(long long a, long long b) { return (a < b) ? a : b; }
const long long N = 1e5 + 10;
long long n, l;
std::vector<vector<long long> > adj(N);
long long timer;
vector<long long> tin(10000), tout(10000);
vector<vector<long long> > up;
vector<long long> lvl(N, 0);
void dfs(long long v, long long p, long long level) {
  tin[v] = ++timer;
  lvl[v] = level;
  up[v][0] = p;
  for (long long i = 1; i < l + 1; i++) {
    up[v][i] = up[up[v][i - 1]][i - 1];
  }
  for (auto i : adj[v])
    if (i != p) dfs(i, v, level + 1);
  tout[v] = ++timer;
}
bool is_ancestor(long long u, long long v) {
  return tin[u] <= tin[v] && tout[u] >= tout[v];
}
long long lca(long long u, long long v) {
  if (is_ancestor(u, v)) return u;
  if (is_ancestor(v, u)) return v;
  for (long long i = l; i >= 0; --i) {
    if (!is_ancestor(up[u][i], v)) u = up[u][i];
  }
  return up[u][0];
}
void preprocess(long long root) {
  tin.resize(n + 1);
  tout.resize(n + 1);
  timer = 0;
  l = ceil(log2(n + 1));
  up.assign(n + 100, vector<long long>(l + 1));
  dfs(root, root, 1);
}
long long dist(long long x, long long y) {
  long long a = lca(x, y);
  return abs(lvl[a] - lvl[x]) + abs(lvl[a] - lvl[y]);
}
long long solve() {
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  preprocess(1);
  long long a = lca(3, 5);
  long long q;
  cin >> q;
  while (q--) {
    long long x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    long long p = dist(a, b);
    if ((p - k) % 2 == 0 && p <= k) {
      cout << "YES" << '\n';
      continue;
    }
    p = dist(a, x) + dist(y, b) + 1;
    if ((p - k) % 2 == 0 && p <= k) {
      cout << "YES" << '\n';
      continue;
    }
    p = dist(a, y) + dist(x, b) + 1;
    if ((p - k) % 2 == 0 && p <= k) {
      cout << "YES" << '\n';
      continue;
    }
    cout << "NO" << '\n';
  }
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
