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 long long mod = 998244353;
int euler[2 * 100005], fir[100005], dep[100005];
int st[8 * 100005];
int cr = 0, n, q;
vector<int> adj[100005];
void dfs(int i, int p) {
  fir[i] = cr;
  euler[cr] = i;
  cr++;
  for (auto x : adj[i]) {
    if (x == p) continue;
    dep[x] = dep[i] + 1;
    dfs(x, i);
    euler[cr] = i;
    cr++;
  }
}
void build(int lo, int hi, int pos) {
  if (lo == hi) {
    st[pos] = euler[lo];
    return;
  }
  int mid = (lo + hi) / 2;
  build(lo, mid, 2 * pos + 1);
  build(mid + 1, hi, 2 * pos + 2);
  if (dep[st[2 * pos + 1]] > dep[st[2 * pos + 2]]) {
    st[pos] = st[2 * pos + 2];
  } else {
    st[pos] = st[2 * pos + 1];
  }
  return;
}
int rq(int qlo, int qhi, int lo, int hi, int pos) {
  if (qlo > hi || qhi < lo) {
    return n;
  }
  if (qlo <= lo && hi <= qhi) {
    return st[pos];
  }
  int mid = (lo + hi) / 2;
  int le = rq(qlo, qhi, lo, mid, 2 * pos + 1);
  int ri = rq(qlo, qhi, mid + 1, hi, 2 * pos + 2);
  if (dep[le] > dep[ri]) {
    return ri;
  } else {
    return le;
  }
}
void swap(int &i, int &j) {
  if (i > j) {
    int tmp = j;
    j = i;
    i = tmp;
  }
}
int lca(int i, int j) {
  i = fir[i];
  j = fir[j];
  swap(i, j);
  return rq(i, j, 0, cr, 0);
}
int dist(int i, int j) {
  int z = lca(i, j);
  return dep[i] + dep[j] - 2 * dep[z];
}
void query() {
  int x, y, a, b, k;
  cin >> x >> y >> a >> b >> k;
  x--;
  y--;
  a--;
  b--;
  int dis[2];
  dis[0] = 2e9;
  dis[1] = 2e9;
  int d1 = dist(a, b);
  dis[d1 % 2] = min(dis[d1 % 2], d1);
  d1 = dist(a, x) + dist(y, b) + 1;
  dis[d1 % 2] = min(dis[d1 % 2], d1);
  d1 = dist(a, y) + dist(x, b) + 1;
  dis[d1 % 2] = min(dis[d1 % 2], d1);
  if (dis[k % 2] <= k) {
    cout << "YES" << '\n';
  } else {
    cout << "NO" << '\n';
  }
  return;
}
void solve() {
  cin >> n;
  for (long long i = (long long)1; i != (long long)n; i = i + 1) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dep[0] = 0;
  dep[n] = 2e9;
  dfs(0, -1);
  build(0, cr, 0);
  cin >> q;
  for (long long i = (long long)0; i != (long long)q; i = i + 1) {
    query();
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  t = 1;
  while (t--) {
    solve();
  }
}
