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>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (T i : v) os << i << ", ";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const set<T> &v) {
  for (T i : v) os << i << " ";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, const multiset<T> &v) {
  for (T i : v) os << i << " ";
  return os;
}
template <class T, class second>
ostream &operator<<(ostream &os, const pair<T, second> &v) {
  os << v.first << ' ' << v.second;
  return os;
}
template <class T, class second>
ostream &operator<<(ostream &os, const map<T, second> &v) {
  for (auto i : v) os << '(' << i.first << " => " << i.second << ')' << ' ';
  return os;
}
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
const int N = 1e5 + 5;
vector<vector<int>> v(N);
vector<int> par(N), level(N), mx(N, 0);
vector<vector<int>> dp(log2(N) + 5, vector<int>(N));
int n;
void dfs_lca(int a, int par, int lev) {
  ::par[a] = par;
  level[a] = lev;
  for (int i : v[a]) {
    if (i == par) continue;
    dfs_lca(i, a, lev + 1);
  }
}
void find_ancestor() {
  for (int i = 1; i <= n; i++) {
    dp[0][i] = par[i];
  }
  for (int i = 1; i < dp.size(); i++) {
    for (int j = 1; j <= n; j++) {
      dp[i][j] = dp[i - 1][dp[i - 1][j]];
    }
  }
}
long long lca(int a, int b) {
  if (level[b] < level[a]) swap(a, b);
  int diff = level[b] - level[a];
  while (diff) {
    int log = log2(diff);
    b = dp[log][b];
    diff -= (1 << log);
  }
  while (a != b) {
    int i = log2(level[a]);
    while (i > 0 && dp[i][a] == dp[i][b]) i--;
    a = dp[i][a];
    b = dp[i][b];
  }
  return a;
}
void compute_lca() {
  dfs_lca(1, 0, 0);
  find_ancestor();
}
int dis(int a, int b) { return level[a] + level[b] - level[lca(a, b)] * 2; }
void solve() {
  compute_lca();
  int q;
  cin >> q;
  while (q--) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    if (level[a] < level[b]) swap(a, b);
    int xy = dis(x, y);
    int t = dis(a, b);
    int t1 = dis(a, x) + dis(b, y) + 1;
    int t4 = dis(a, y) + dis(b, x) + 1;
    if (t <= k && (k - t) % 2 == 0) {
      cout << "YES\n";
      continue;
    }
    if (t1 <= k && (k - t1) % 2 == 0) {
      cout << "YES\n";
      continue;
    }
    if (t4 <= k && (k - t4) % 2 == 0) {
      cout << "YES\n";
      continue;
    }
    cout << "NO\n";
  }
}
int main() {
  fast();
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  solve();
}
