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;
const int mxn = 1e5;
vector<int> adj[mxn + 1];
bool vis[mxn + 1];
int dep[mxn + 1], first[mxn + 1];
int n;
vector<int> euler, t;
void dfs(int v, int d = 0) {
  vis[v] = true;
  dep[v] = d;
  first[v] = euler.size();
  euler.push_back(v);
  for (int nbr : adj[v]) {
    if (!vis[nbr]) {
      dfs(nbr, d + 1);
      euler.push_back(v);
    }
  }
}
int minh(int u, int v) {
  if (dep[u] < dep[v])
    return u;
  else
    return v;
}
void build() {
  for (int i = euler.size() - 1; i > 0; --i)
    t[i] = minh(t[i << 1], t[i << 1 | 1]);
}
int query(int l, int r) {
  int out = 0;
  for (l += euler.size(), r += euler.size(); l < r; l >>= 1, r >>= 1) {
    if (l & 1) out = minh(out, t[l++]);
    if (r & 1) out = minh(out, t[--r]);
  }
  return out;
}
int lca(int u, int v) {
  int l = min(first[u], first[v]);
  int r = max(first[u], first[v]);
  return query(l, r + 1);
}
int dist(int u, int v) {
  int a = lca(u, v);
  return dep[u] + dep[v] - 2 * dep[a];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1);
  dep[0] = INT_MAX;
  t.resize(2 * euler.size());
  for (int i = 0; i < euler.size(); ++i) t[i + euler.size()] = euler[i];
  build();
  int q;
  cin >> q;
  while (q--) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    int ab = dist(a, b);
    if (ab <= k && ab % 2 == k % 2) {
      cout << "YES\n";
      continue;
    }
    int via = min(dist(a, x) + dist(b, y) + 1, dist(a, y) + dist(b, x) + 1);
    if (via <= k && via % 2 == k % 2) {
      cout << "YES\n";
    } else {
      cout << "NO\n";
    }
  }
}
