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;
template <class T>
bool uin(T& a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool uax(T& a, T b) {
  return a < b ? (a = b, true) : false;
}
const int nax = 1 << 17;
const int LOG = 17;
vector<int> G[nax];
int N, Q;
int in[nax], out[nax], dep[nax];
int jmp[nax][LOG + 1];
int TIME;
void dfs(int v, int p) {
  in[v] = ++TIME;
  jmp[v][0] = p;
  for (int i = 1; i <= LOG; ++i) jmp[v][i] = jmp[jmp[v][i - 1]][i - 1];
  for (int ch : G[v]) {
    if (ch == p) continue;
    dep[ch] = dep[v] + 1;
    dfs(ch, v);
  }
  out[v] = ++TIME;
}
inline bool is_ancestor(int u, int v) {
  return in[u] <= in[v] && out[u] >= out[v];
}
int lca(int u, int v) {
  if (is_ancestor(u, v)) return u;
  if (is_ancestor(v, u)) return v;
  for (int i = LOG; i >= 0; --i)
    if (!is_ancestor(jmp[u][i], v)) u = jmp[u][i];
  return jmp[u][0];
}
inline int dist(int u, int v) { return dep[u] + dep[v] - 2 * dep[lca(u, v)]; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N;
  for (int i = 0; i < N - 1; ++i) {
    int u, v;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 1);
  cin >> Q;
  for (int tt = 1; tt <= Q; ++tt) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    int bef = dist(a, b);
    int aft = min(dist(a, x) + dist(y, b), dist(a, y) + dist(x, b)) + 1;
    int valid_dist = 1e9 + 10;
    if (bef % 2 == k % 2) uin(valid_dist, bef);
    if (aft % 2 == k % 2) uin(valid_dist, aft);
    cout << (valid_dist <= k ? "YES" : "NO") << '\n';
  }
}
