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;
int n, a, b, q, x, y, k;
vector<int> G[100009];
int vis[100009], dist[100009];
int p[100009][20], t[100009];
void DFS(int v) {
  vis[v] = 1;
  for (int i = 0; i < G[v].size(); i++) {
    int w = G[v][i];
    if (vis[w] == 0) {
      t[w] = v;
      dist[w] = dist[v] + 1;
      DFS(w);
    }
  }
}
void make_table() {
  for (int i = 1; i <= n; i++) {
    for (int j = 0; (1 << j) < n; j++) {
      p[i][j] = -1;
    }
  }
  for (int i = 2; i <= n; i++) {
    p[i][0] = t[i];
  }
  for (int j = 1; (1 << j) < n; j++) {
    for (int i = 1; i <= n; i++) {
      if (p[i][j - 1] != -1) {
        p[i][j] = p[p[i][j - 1]][j - 1];
      }
    }
  }
}
int lca(int v, int w) {
  if (dist[v] < dist[w]) {
    swap(v, w);
  }
  int log;
  for (log = 1; (1 << log) < dist[v]; log++)
    ;
  for (int i = log; i >= 0; i--) {
    if (dist[v] - (1 << i) >= dist[w]) v = p[v][i];
  }
  if (v == w) return v;
  for (int i = log; i >= 0; i--) {
    if (p[v][i] != -1 && p[v][i] != p[w][i]) {
      v = p[v][i];
      w = p[w][i];
    }
  }
  return t[v];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  DFS(1);
  make_table();
  cin >> q;
  while (q--) {
    cin >> x >> y >> a >> b >> k;
    int len = dist[a] + dist[b] - 2 * dist[lca(a, b)];
    int mod = abs((dist[a] % 2) - (dist[b] % 2));
    if (k % 2 == mod && len <= k) {
      cout << "YES\n";
      continue;
    }
    len = dist[a] + dist[x] - 2 * dist[lca(a, x)] + 1 + dist[y] + dist[b] -
          2 * dist[lca(y, b)];
    mod = abs((dist[a] % 2) - (dist[x] % 2)) +
          abs((dist[y] % 2) - (dist[b] % 2)) + 1;
    mod %= 2;
    if (k % 2 == mod && len <= k) {
      cout << "YES\n";
      continue;
    }
    len = dist[a] + dist[y] - 2 * dist[lca(a, y)] + 1 + dist[x] + dist[b] -
          2 * dist[lca(x, b)];
    mod = abs((dist[a] % 2) - (dist[y] % 2)) +
          abs((dist[x] % 2) - (dist[b] % 2)) + 1;
    mod %= 2;
    if (k % 2 == mod && len <= k) {
      cout << "YES\n";
      continue;
    }
    cout << "NO\n";
  }
}
