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 MOD = 1e9 + 7;
const long long INF = 1e18;
const double EPS = 1e-6;
const int MAX_N = 3e5 + 5;
const int MAX_LOG = 25;
int N;
vector<int> adjList[MAX_N];
int depth[MAX_N], parent[MAX_N][MAX_LOG];
void DFS(int u, int p, int d) {
  depth[u] = d;
  parent[u][0] = p;
  for (int v : adjList[u]) {
    if (v != p) DFS(v, u, d + 1);
  }
}
void LCA_precompute() {
  DFS(1, 0, 0);
  for (int k = 1; k < MAX_LOG; ++k) {
    for (int i = 1; i <= N; ++i) {
      parent[i][k] = parent[parent[i][k - 1]][k - 1];
    }
  }
}
int lift(int u, int h) {
  for (int i = 0; (1 << i) <= h; ++i) {
    if ((h & (1 << i)) != 0) {
      u = parent[u][i];
    }
  }
  return u;
}
int LCA(int a, int b) {
  if (depth[a] < depth[b]) {
    b = lift(b, depth[b] - depth[a]);
  } else if (depth[b] < depth[a]) {
    a = lift(a, depth[a] - depth[b]);
  }
  if (a == b) return a;
  for (int k = MAX_LOG - 1; k >= 0; --k) {
    if (parent[a][k] != parent[b][k]) {
      a = parent[a][k], b = parent[b][k];
    }
  }
  return parent[a][0];
}
int dist(int a, int b) {
  int lca = LCA(a, b);
  return depth[a] + depth[b] - 2 * depth[lca];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N;
  for (int i = 0; i < N - 1; ++i) {
    int u, v;
    cin >> u >> v;
    adjList[u].push_back(v);
    adjList[v].push_back(u);
  }
  LCA_precompute();
  int q;
  cin >> q;
  while (q-- > 0) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    bool ans = false;
    if (dist(a, b) <= k && (k - dist(a, b)) % 2 == 0) ans = true;
    int d = dist(a, x) + dist(b, y) + 1;
    if (d <= k && (k - d) % 2 == 0) ans = true;
    d = dist(b, x) + dist(a, y) + 1;
    if (d <= k && (k - d) % 2 == 0) ans = true;
    if (ans)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
