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 mod = 1e9 + 7;
inline long long add(long long a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
  return a;
}
inline long long sub(long long a, long long b) {
  a -= b;
  if (a < 0) a += mod;
  return a;
}
inline long long mul(long long a, long long b) {
  return (long long)((long long)a * b % mod);
}
vector<vector<long long> > adj, dp;
vector<long long> cnt, lvl;
void DFSUtil(long long u, long long p) {
  if (u != 0) {
    lvl[u] = lvl[p] + 1;
  }
  dp[u][0] = p;
  for (long long i = (1); i <= (20); i++) {
    dp[u][i] = dp[dp[u][i - 1]][i - 1];
  }
  for (auto it : adj[u])
    if (it != p) DFSUtil(it, u);
}
void DFS() {
  long long V = adj.size();
  lvl.assign(V, 0);
  DFSUtil(0, 0);
}
long long lca(long long x, long long y) {
  if (x == y) return 0;
  if (lvl[x] < lvl[y]) swap(x, y);
  long long d = lvl[x] - lvl[y];
  long long x1 = x;
  for (long long i = (0); i <= (20); i++)
    if ((1 << i) & d) x1 = dp[x1][i];
  if (x1 == y) return d;
  long long xx = x1, yy = y;
  for (long long i = (20); i >= (0); i--)
    if (dp[xx][i] != dp[yy][i]) {
      d += 2 * (1 << i);
      xx = dp[xx][i];
      yy = dp[yy][i];
    }
  d += 2;
  return d;
}
bool query() {
  long long a, b, x, y, k;
  cin >> a >> b >> x >> y >> k;
  x--;
  y--;
  a--;
  b--;
  long long v1 = lca(x, y), v2 = lca(x, a), v3 = lca(x, b), v4 = lca(y, a),
            v5 = lca(y, b);
  if (v1 <= k && (k - v1) % 2 == 0) {
    return true;
  }
  if ((v2 + v5 + 1) <= k && (k - (v2 + v5 + 1)) % 2 == 0) {
    return true;
  }
  if ((v3 + v4 + 1) <= k && (k - (v3 + v4 + 1)) % 2 == 0) {
    return true;
  }
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  while (t--) {
    long long n;
    cin >> n;
    adj.resize(n + 1);
    for (long long i = (1); i <= (n - 1); i++) {
      long long p, q;
      cin >> p >> q;
      p--;
      q--;
      adj[p].push_back(q);
      adj[q].push_back(p);
    }
    dp.assign(n + 1, vector<long long>(21, 0));
    DFS();
    long long m;
    cin >> m;
    for (long long i = (1); i <= (m); i++) {
      if (query())
        cout << "YES\n";
      else
        cout << "NO\n";
    }
  }
  return 0;
}
