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;
long long level[1 << 18];
long long dp[1 << 18][18];
long long n;
vector<vector<long long>> g;
void dfs(long long u, long long par) {
  if (par == -1)
    level[u] = 0;
  else
    level[u] = level[par] + 1;
  dp[u][0] = par;
  for (auto it : g[u]) {
    if (it == par) continue;
    dfs(it, u);
  }
}
void build_table() {
  long long max_jumps = ceil(log2(n));
  for (long long i = 0; i < max_jumps - 1; i++) {
    for (long long j = 1; j <= n; j++) {
      if (dp[j][i] != -1) dp[j][i + 1] = dp[dp[j][i]][i];
    }
  }
}
long long LCA(long long u, long long v) {
  long long max_jumps = ceil(log2(n));
  if (level[u] < level[v]) swap(u, v);
  long long diff = level[u] - level[v];
  for (long long i = 0; i < max_jumps; i++) {
    if (diff & (1 << i)) u = dp[u][i];
  }
  if (u == v) return u;
  for (long long i = max_jumps - 1; i >= 0; i--) {
    if (dp[u][i] != dp[v][i]) {
      u = dp[u][i];
      v = dp[v][i];
    }
  }
  return dp[u][0];
}
long long dist(long long u, long long v) {
  return (level[u] + level[v] - 2 * level[LCA(u, v)]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long q, u, v, x, y, a, b, k;
  cin >> n;
  g.resize(n + 1);
  for (long long i = 0; i < n - 1; i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  memset(dp, -1, sizeof(dp));
  dfs(1, -1);
  build_table();
  cin >> q;
  while (q--) {
    cin >> x >> y >> a >> b >> k;
    pair<long long, long long> p = {0x3f3f3f3f, 0x3f3f3f3f};
    long long d = dist(a, b);
    if (d % 2)
      p.second = d;
    else
      p.first = d;
    d = dist(a, x) + 1 + dist(y, b);
    if (d % 2)
      p.second = min(d, p.second);
    else
      p.first = min(d, p.first);
    d = dist(a, y) + 1 + dist(x, b);
    if (d % 2)
      p.second = min(d, p.second);
    else
      p.first = min(d, p.first);
    bool flag = (k % 2) ? (p.second != 0x3f3f3f3f && k >= p.second)
                        : (p.first != 0x3f3f3f3f && k >= p.first);
    flag ? cout << "YES\n" : cout << "NO\n";
  }
  return 0;
}
