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;
vector<vector<int>> edge;
vector<int> dist;
vector<int> et;
vector<int> first;
void inedge(int n, int m) {
  edge.resize(n + 1);
  dist.resize(n + 1, -1);
  first.resize(n + 1);
  int a, b;
  for (int i = 0; i < m; i++) {
    cin >> a >> b;
    edge[a].push_back(b);
    edge[b].push_back(a);
  }
}
void dfs(int v, int p, int h) {
  if (dist[v] != -1) return;
  dist[v] = h;
  first[v] = et.size();
  et.push_back(v);
  for (int i = 0; i < edge[v].size(); i++) {
    if (p - edge[v][i]) {
      dfs(edge[v][i], v, h + 1);
      et.push_back(v);
    }
  }
}
vector<int> t;
int n;
void build(vector<int>& vv) {
  n = vv.size();
  t.resize(2 * n, INT_MAX);
  for (int i = 0; i < n; i++) t[i + n] = vv[i];
  for (int i = n - 1; i > 0; i--) {
    t[i] = min(t[i << 1], t[i << 1 | 1]);
  }
}
int rq(int l, int r) {
  int res = INT_MAX;
  r++;
  for (l += n, r += n; l < r; l >>= 1, r >>= 1) {
    if (l & 1) res = min(res, t[l++]);
    if (r & 1) res = min(res, t[--r]);
  }
  return res;
}
int dt(int a, int b) {
  if (first[a] > first[b]) swap(a, b);
  int dlca = rq(first[a], first[b]);
  return dist[a] + dist[b] - 2 * dlca;
}
int main() {
  int n;
  cin >> n;
  inedge(n, n - 1);
  dfs(1, -1, 0);
  vector<int> vv;
  for (int i = 0; i < et.size(); i++) vv.push_back(dist[et[i]]);
  build(vv);
  int q, x, y, a, b, k;
  cin >> q;
  while (q--) {
    cin >> x >> y >> a >> b >> k;
    int d1 = dt(a, b);
    int d2 = dt(a, x) + dt(y, b) + 1;
    int d3 = dt(a, y) + dt(x, b) + 1;
    if (d1 <= k && (d1 % 2 == k % 2)) {
      cout << "YES\n";
    } else if (d2 <= k && (d2 % 2 == k % 2)) {
      cout << "YES\n";
    } else if (d3 <= k && (d3 % 2 == k % 2)) {
      cout << "YES\n";
    } else
      cout << "NO\n";
  }
}
