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;
vector<long long int> parents(vector<vector<long long int>> &G, int n) {
  vector<int> Vis(n + 1, 0);
  vector<long long int> P(n + 1, 0);
  P[1] = -1;
  Vis[1] = 1;
  queue<long long int> Q;
  Q.push(1);
  while (!Q.empty()) {
    int t = Q.front();
    Q.pop();
    for (int i = 0; i < G[t].size(); i++) {
      int kaim = G[t][i];
      if (Vis[kaim] == 0) {
        P[kaim] = t;
        Q.push(kaim);
        Vis[kaim] = 1;
      }
    }
  }
  return P;
}
long long int lca(long long int a, long long int b,
                  vector<vector<long long int>> &P2, vector<long long int> &D) {
  if (a == 1 || b == 1) {
    return 1;
  }
  if (D[a] > D[b]) {
    swap(a, b);
  }
  long long int x = D[b] - D[a];
  int i = 0;
  while (x > 0) {
    if (x % 2 == 1) {
      b = P2[b][i];
    }
    i++;
    x = x / 2;
  }
  if (a == b) {
    return a;
  }
  for (int i = 19; i >= 0; i--) {
    long long int a1 = P2[a][i];
    long long int b1 = P2[b][i];
    if (a1 != b1 && a1 > 0 && b1 > 0) {
      a = a1;
      b = b1;
    }
  }
  return P2[a][0];
}
vector<long long int> atstumai(vector<vector<long long int>> &G, int a) {
  vector<long long int> D(G.size(), -1);
  queue<long long int> Q;
  Q.push(a);
  D[a] = 0;
  while (!Q.empty()) {
    int t = Q.front();
    Q.pop();
    for (int i = 0; i < G[t].size(); i++) {
      int kaim = G[t][i];
      if (D[kaim] == -1) {
        D[kaim] = D[t] + 1;
        Q.push(kaim);
      }
    }
  }
  return D;
}
long long int dist(long long int a, long long int b,
                   vector<vector<long long int>> &P2,
                   vector<long long int> &D) {
  return D[a] + D[b] - 2 * D[lca(a, b, P2, D)];
}
void solve(long long int x, long long int y, long long int a, long long int b,
           long long int k, vector<vector<long long int>> &P2,
           vector<long long int> &D) {
  long long int v = dist(a, b, P2, D);
  if (v <= k && (k - v) % 2 == 0) {
    cout << "YES" << endl;
    return;
  }
  long long int a2 = dist(a, x, P2, D) + dist(b, y, P2, D) + 1;
  if (a2 <= k && (k - a2) % 2 == 0) {
    cout << "YES" << endl;
    return;
  }
  long long int a3 = dist(a, y, P2, D) + dist(b, x, P2, D) + 1;
  if (a3 <= k && (k - a3) % 2 == 0) {
    cout << "YES" << endl;
    return;
  }
  cout << "NO" << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<vector<long long int>> G(n + 1);
  for (int i = 0; i < n - 1; i++) {
    long long int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  vector<long long int> D = atstumai(G, 1);
  vector<long long int> P = parents(G, n);
  vector<vector<long long int>> P2(n + 1);
  for (int i = 0; i < n + 1; i++) {
    P2[i].push_back(P[i]);
  }
  for (int i = 1; i < 20; i++) {
    for (int j = 0; j < n + 1; j++) {
      long long int x = P2[j][i - 1];
      if (x == -1) {
        P2[j].push_back(-1);
      } else {
        P2[j].push_back(P2[x][i - 1]);
      }
    }
  }
  int q;
  cin >> q;
  for (int f = 0; f < q; f++) {
    long long int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    solve(x, y, a, b, k, P2, D);
  }
  return 0;
}
