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 int n, l, timer = 0;
vector<long long int> adj[100005];
vector<vector<long long int> > up(100005, vector<long long int>(20));
vector<long long int> tin(100005), tout(100005), dist(100005, 0);
void dfs(long long int v, long long int p) {
  tin[v] = ++timer;
  up[v][0] = p;
  for (long long int i = 1; i <= l; ++i) up[v][i] = up[up[v][i - 1]][i - 1];
  for (long long int u : adj[v]) {
    if (u != p) {
      dist[u] = dist[v] + 1LL;
      dfs(u, v);
    }
  }
  tout[v] = ++timer;
}
bool is_ancestor(long long int u, long long int v) {
  return ((tin[u] <= tin[v]) && (tout[u] >= tout[v]));
}
long long int lca(long long int u, long long int v) {
  if (is_ancestor(u, v)) return u;
  if (is_ancestor(v, u)) return v;
  for (long long int i = l; i >= 0; i--) {
    if (!is_ancestor(up[u][i], v)) u = up[u][i];
  }
  return up[u][0];
}
long long int length(long long int u, long long int v) {
  long long int req = lca(u, v);
  return (dist[u] + dist[v] - 2 * dist[req]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (long long int i = 0; i <= n - 2; i++) {
    long long int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  timer = 0;
  l = ceil(log2(n));
  long long int root = 1;
  dfs(root, root);
  long long int w;
  cin >> w;
  while (w--) {
    long long int x, y, a, b, k;
    cin >> a >> b >> x >> y >> k;
    long long int dis = length(x, y);
    if (dis <= k && ((dis % 2) == (k % 2))) {
      cout << "YES" << '\n';
      continue;
    }
    dis = length(x, a) + length(y, b) + 1LL;
    if (dis <= k && ((dis % 2) == (k % 2))) {
      cout << "YES" << '\n';
      continue;
    }
    dis = length(x, b) + length(y, a) + 1LL;
    if (dis <= k && ((dis % 2) == (k % 2))) {
      cout << "YES" << '\n';
      continue;
    }
    cout << "NO" << '\n';
  }
  return 0;
}
