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;
struct LCA {
  vector<int> height, euler, first, segtree, parent;
  vector<bool> visited;
  int n;
  LCA(vector<vector<int> > &adj, int root = 0) {
    n = adj.size();
    height.resize(n);
    first.resize(n);
    parent.resize(n);
    euler.reserve(n * 2);
    visited.assign(n, false);
    parent[root] = -1;
    dfs(adj, root);
    int m = euler.size();
    segtree.resize(m * 4);
    build(1, 0, m - 1);
  }
  void dfs(vector<vector<int> > &adj, int node, int h = 0) {
    visited[node] = true;
    height[node] = h;
    first[node] = euler.size();
    euler.push_back(node);
    for (auto to : adj[node]) {
      if (!visited[to]) {
        parent[to] = node;
        dfs(adj, to, h + 1);
        euler.push_back(node);
      }
    }
  }
  void build(int node, int b, int e) {
    if (b == e) {
      segtree[node] = euler[b];
    } else {
      int mid = (b + e) / 2;
      build(node << 1, b, mid);
      build(node << 1 | 1, mid + 1, e);
      int l = segtree[node << 1], r = segtree[node << 1 | 1];
      segtree[node] = (height[l] < height[r]) ? l : r;
    }
  }
  int query(int node, int b, int e, int L, int R) {
    if (b > R || e < L) return -1;
    if (b >= L && e <= R) return segtree[node];
    int mid = (b + e) >> 1;
    int left = query(node << 1, b, mid, L, R);
    int right = query(node << 1 | 1, mid + 1, e, L, R);
    if (left == -1) return right;
    if (right == -1) return left;
    return height[left] < height[right] ? left : right;
  }
  int lca(int u, int v) {
    int left = first[u], right = first[v];
    if (left > right) swap(left, right);
    return query(1, 0, euler.size() - 1, left, right);
  }
};
void solve() {
  int n;
  cin >> n;
  vector<vector<int> > adj(n);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    x--, y--;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  LCA Tree(adj, 0);
  auto dis = [&](int u, int v) {
    int lc = Tree.lca(u, v);
    return Tree.height[u] + Tree.height[v] - 2 * Tree.height[lc];
  };
  int q;
  cin >> q;
  while (q--) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    x--, y--, a--, b--;
    bool ok = false;
    vector<int> pos;
    int ab = dis(a, b);
    int ax = dis(a, x);
    int ay = dis(a, y);
    int bx = dis(b, x);
    int by = dis(b, y);
    pos.push_back(ab);
    pos.push_back(ax + 1 + by);
    pos.push_back(ax + 1 + ay + ab);
    pos.push_back(ay + 1 + bx);
    ;
    for (auto &d : pos) {
      if (d <= k) {
        ok |= ((d & 1) == (k & 1));
      }
    }
    if (ok) {
      cout << "YES\n";
    } else {
      cout << "NO\n";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
}
