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 <typename T>
class segtree {
 private:
  int n, sz;
  vector<pair<T, int> > node;

 public:
  void resize(vector<T>& v) {
    sz = (int)v.size();
    n = 1;
    while (n < sz) {
      n *= 2;
    }
    node.resize(2 * n);
    for (int i = 0; i < sz; i++) {
      node[i + n] = make_pair(v[i], i);
    }
    for (int i = n - 1; i >= 1; i--) {
      node[i] = min(node[2 * i], node[2 * i + 1]);
    }
  }
  void update(int k, T a) {
    node[k += n] = make_pair(a, k);
    while (k >>= 1) {
      node[k] = min(node[2 * k], node[2 * k + 1]);
    }
  }
  pair<T, int> query(int a, int b) {
    pair<T, int> res1 = make_pair(numeric_limits<T>::max(), -1);
    pair<T, int> res2 = make_pair(numeric_limits<T>::max(), -1);
    a += n, b += n;
    while (a != b) {
      if (a % 2) res1 = min(res1, node[a++]);
      if (b % 2) res2 = min(res2, node[--b]);
      a >>= 1, b >>= 1;
    }
    return min(res1, res2);
  }
};
class LCA {
 private:
  int V;
  vector<vector<int> > G;
  vector<int> ord, depth, id;
  segtree<int> st;
  void dfs(int u, int p, int k) {
    id[u] = (int)ord.size();
    ord.push_back(u);
    depth[u] = k;
    for (int v : G[u]) {
      if (v != p) {
        dfs(v, u, k + 1);
        ord.push_back(u);
      }
    }
  }

 public:
  LCA(int node_size) : V(node_size), G(V), depth(V), id(V, -1) {}
  void add_edge(int from, int to) {
    G[from].push_back(to), G[to].push_back(from);
  }
  void build() {
    ord.reserve(2 * V - 2);
    for (int i = 0; i < V; i++) {
      if (id[i] < 0) {
        dfs(i, -1, 0);
      }
    }
    vector<int> stvec(2 * V - 2);
    for (int i = 0; i < 2 * V - 2; i++) {
      stvec[i] = depth[ord[i]];
    }
    st.resize(stvec);
  }
  int lca(int u, int v) {
    return ord[st.query(min(id[u], id[v]), max(id[u], id[v]) + 1).second];
  }
  int dist(int u, int v) {
    int lca_ = lca(u, v);
    return depth[u] + depth[v] - 2 * depth[lca_];
  }
};
void solve();
int main(void) {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int t = 1;
  for (int _ = 0; _ < t; _++) solve();
  return 0;
}
void solve() {
  int n;
  cin >> n;
  LCA G(n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    G.add_edge(a, b);
  }
  G.build();
  int q;
  cin >> q;
  for (int _ = 0; _ < q; _++) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    x--, y--, a--, b--;
    int d = G.dist(a, b);
    if (k >= d and (k - d) % 2 == 0) {
      cout << "YES" << '\n';
    } else {
      int d2 = min(G.dist(a, x) + G.dist(y, b), G.dist(a, y) + G.dist(x, b));
      d2++;
      if (k >= d2 and (k - d2) % 2 == 0) {
        cout << "YES" << '\n';
      } else {
        cout << "NO" << '\n';
      }
    }
  }
  return;
}
