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 long long N = 1e5 + 5;
const long long L = ceil(log2(N)) + 1;
vector<vector<long long> > adj(N, vector<long long int>());
vector<long long int> tin(N, 0), tout(N, 0);
vector<vector<long long> > up(N, vector<long long int>(L));
vector<long long int> d(N, 0);
long long n;
long long tr = 0;
void dfs(long long u, long long p) {
  tin[u] = tr++;
  if (p != -1) {
    d[u] = d[p] + 1;
    for (long long int i = 0; i < L; i++) {
      if (i == 0)
        up[u][0] = p;
      else
        up[u][i] = up[up[u][i - 1]][i - 1];
    }
  }
  for (long long int i = 0; i < (long long int)adj[u].size(); i++)
    if (adj[u][i] != p) dfs(adj[u][i], u);
  tout[u] = tr++;
  return;
}
long long 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 - 1; i >= 0; i--) {
    if (is_ancestor(up[u][i], v) == 0) u = up[u][i];
  }
  return up[u][0];
}
long long dist(long long u, long long v) {
  long long an = lca(u, v);
  return d[u] - d[an] + d[v] - d[an];
}
int solve() {
  cin >> n;
  for (long long int i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    x--, y--;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(0, -1);
  long long q;
  cin >> q;
  while (q--) {
    long long x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    x--, y--, a--, b--;
    long long d1 = dist(a, b);
    long long d2 = dist(a, x) + 1 + dist(y, b);
    long long d3 = dist(a, y) + 1 + dist(x, b);
    if ((d1 <= k && (d1 - k) % 2 == 0) || (d2 <= k && (d2 - k) % 2 == 0) ||
        (d3 <= k && (d3 - k) % 2 == 0))
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
int main() {
  auto start = chrono::high_resolution_clock::now();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long test_cases = 1;
  while (test_cases--) solve();
  auto stop = chrono::high_resolution_clock::now();
  auto duration = chrono::duration_cast<chrono::milliseconds>(stop - start);
}
