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 T>
void __f(const char*, T&&);
template <typename T1, typename... T2>
void __f(const char*, T1&&, T2&&...);
const int N = 2e5 + 5;
const int maxHeight = 20 + 1;
int level[N], parent[N], dp[N][maxHeight];
vector<vector<int>> gph(N);
int n, q;
void dfs(int u, int par, int lvl) {
  parent[u] = par;
  level[u] = lvl;
  for (auto v : gph[u])
    if (v != par) dfs(v, u, lvl + 1);
}
void fill_dp() {
  for (int i = 1; i <= n; i++) dp[i][0] = parent[i];
  for (int j = 1; j < maxHeight; j++) {
    for (int i = 1; i <= n; i++) {
      dp[i][j] = dp[dp[i][j - 1]][j - 1];
    }
  }
}
int lca(int u, int v) {
  if (level[v] < level[u]) {
    swap(u, v);
  }
  int difference = level[v] - level[u];
  while (difference > 0) {
    int _log = log2(difference);
    v = dp[v][_log];
    difference -= (1 << _log);
  }
  while (u != v) {
    int i = log2(level[u]);
    while (i > 0 && dp[u][i] == dp[v][i]) i--;
    u = dp[u][i];
    v = dp[v][i];
  }
  return u;
}
long long distance_btw_two(int u, int v) {
  return level[u] + level[v] - 2 * level[lca(u, v)];
}
void solve() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    gph[u].push_back(v);
    gph[v].push_back(u);
  }
  dfs(1, 0, 0);
  fill_dp();
  cin >> q;
  while (q--) {
    long long x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    long long dist_a_b = 0, min_dist = 0, ans = INT32_MAX;
    dist_a_b = distance_btw_two(a, b);
    min_dist = min(distance_btw_two(a, x) + distance_btw_two(b, y),
                   distance_btw_two(a, y) + distance_btw_two(b, x)) +
               1;
    if (k % 2 == dist_a_b % 2) ans = dist_a_b;
    if (k % 2 == min_dist % 2) ans = min(ans, min_dist);
    if (ans <= k)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
template <typename T>
void __f(const char* variable_name, T&& value) {
  cout << variable_name << ": " << value << endl;
}
template <typename T1, typename... T2>
void __f(const char* names, T1&& arg1, T2&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}
