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>> graf;
vector<int> timein;
vector<int> timeout;
int timer = 0;
const int kolvo = 2e5;
int binup[kolvo][19];
int a, l;
int rast[kolvo];
void dfs(int v, int deep, int p = 0) {
  rast[v] = deep;
  timein[v] = timer++;
  binup[v][0] = p;
  for (int q = 1; q <= l; q++) {
    binup[v][q] = binup[binup[v][q - 1]][q - 1];
  }
  for (int q = 0; q < graf[v].size(); q++) {
    if (graf[v][q] != p) {
      dfs(graf[v][q], deep + 1, v);
    }
  }
  timeout[v] = timer++;
}
bool check(int v1, int v2) {
  return timein[v1] <= timein[v2] && timeout[v1] >= timeout[v2];
}
int lca(int v1, int v2) {
  if (check(v1, v2)) {
    return v1;
  }
  if (check(v2, v1)) {
    return v2;
  }
  for (int q = l; q >= 0; q--) {
    if (!check(binup[v1][q], v2)) {
      v1 = binup[v1][q];
    }
  }
  return binup[v1][0];
}
int rrast(int a, int b) { return rast[a] + rast[b] - 2 * rast[lca(a, b)]; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  rast[0] = 0;
  int b, c, t;
  cin >> a;
  graf.resize(a);
  for (int q = 1; q < a; q++) {
    cin >> b >> c;
    graf[c - 1].push_back(b - 1);
    graf[b - 1].push_back(c - 1);
  }
  timeout.resize(a);
  timein.resize(a);
  while ((1 << l) < a) {
    l++;
  }
  dfs(0, 0, 0);
  cin >> t;
  int d, e, f, g;
  while (t--) {
    cin >> c >> d >> e >> f >> g;
    if (rrast(e - 1, f - 1) % 2 == g % 2 && rrast(e - 1, f - 1) <= g) {
      cout << "YES" << endl;
    } else if ((rrast(e - 1, c - 1) + rrast(f - 1, d - 1) + 1) % 2 == g % 2 &&
               (rrast(e - 1, c - 1) + rrast(f - 1, d - 1) + 1) <= g) {
      cout << "YES" << endl;
    } else if ((rrast(f - 1, c - 1) + rrast(e - 1, d - 1) + 1) % 2 == g % 2 &&
               (rrast(f - 1, c - 1) + rrast(e - 1, d - 1) + 1) <= g) {
      cout << "YES" << endl;
    } else {
      cout << "NO" << endl;
    }
  }
}
