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 maxn = 2e5 + 10;
const int lg = 30;
int n, m;
int h[maxn];
bool mark[maxn];
int f[lg][maxn];
string ans[maxn];
vector<int> adj[maxn];
void in();
void dfs(int u);
void solve();
int dis(int u, int v);
void out();
int main() {
  in();
  solve();
  out();
}
void in() {
  ios::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  cin >> m;
}
void dfs(int u) {
  for (int i = 1; i < lg; i++)
    if (f[i - 1][u] != -1) f[i][u] = f[i - 1][f[i - 1][u]];
  mark[u] = 1;
  for (auto v : adj[u]) {
    if (!mark[v]) {
      h[v] = h[u] + 1;
      f[0][v] = u;
      dfs(v);
    }
  }
}
void solve() {
  for (int i = 0; i < lg; i++) fill(f[i], f[i] + n, -1);
  dfs(0);
  for (int t = 0; t < m; t++) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    x--;
    y--;
    a--;
    b--;
    ans[t] = "NO\n";
    int ab = dis(a, b);
    int xy = dis(x, y);
    int ax = dis(a, x);
    int ay = dis(a, y);
    int bx = dis(b, x);
    int by = dis(b, y);
    if (ab <= k && (k - ab) % 2 == 0) ans[t] = "YES\n";
    ab = ax + 1 + by;
    if (ab <= k && (k - ab) % 2 == 0) ans[t] = "YES\n";
    ab = ay + 1 + bx;
    if (ab <= k && (k - ab) % 2 == 0) ans[t] = "YES\n";
  }
}
int dis(int u, int v) {
  int tmp = h[u] + h[v];
  if (h[u] > h[v]) swap(u, v);
  for (int i = lg - 1; i >= 0; i--)
    if (f[i][v] != -1 && h[f[i][v]] >= h[u]) v = f[i][v];
  if (u == v) return tmp - 2 * h[u];
  for (int i = lg - 1; i >= 0; i--) {
    if (f[i][v] != f[i][u]) {
      v = f[i][v];
      u = f[i][u];
    }
  }
  u = f[0][u];
  return tmp - 2 * h[u];
}
void out() {
  for (int i = 0; i < m; i++) cout << ans[i];
}
