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 MAX_N = 100000;
const int LIM = 17;
const int INF = (int)1e9 + 7;
vector<int> adj[MAX_N + 5];
int depth[MAX_N + 5];
int par[MAX_N + 5][LIM + 1];
void build(int cur, int p) {
  int i;
  depth[cur] = depth[p] + 1;
  par[cur][0] = p;
  for (i = 1; i <= LIM; i++) par[cur][i] = par[par[cur][i - 1]][i - 1];
  for (int x : adj[cur])
    if (x != p) build(x, cur);
}
int lca_len(int a, int b) {
  int i, len = 0;
  if (depth[a] > depth[b]) swap(a, b);
  for (i = LIM; i >= 0; i--) {
    if (depth[par[b][i]] >= depth[a]) {
      b = par[b][i];
      len += (1 << i);
    }
  }
  if (a == b) return len;
  for (i = LIM; i >= 0; i--) {
    if (par[a][i] != par[b][i]) {
      a = par[a][i];
      b = par[b][i];
      len += (1 << (i + 1));
    }
  }
  return len + 2;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, q, i;
  cin >> n;
  for (i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  build(1, 0);
  cin >> q;
  while (q--) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    int without = lca_len(a, b);
    int with =
        min(lca_len(a, x) + lca_len(y, b), lca_len(a, y) + lca_len(x, b)) + 1;
    int ans = INF;
    if (without % 2 == k % 2) ans = without;
    if (with % 2 == k % 2) ans = min(ans, with);
    cout << (ans <= k ? "YES" : "NO") << '\n';
  }
}
