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>
long long R = 7 + 1e9, NUMTESTCASE;
const int NN = 10 + 2e6;
const double pi = acos(-1.0);
int di[4] = {1, 0, -1, 0}, dj[4] = {0, 1, 0, -1},
    DI[8] = {1, 1, 0, -1, -1, -1, 0, 1}, DJ[8] = {0, 1, 1, 1, 0, -1, -1, -1};
using namespace std;
int n, ti = 1, timein[NN], timeout[NN], lev[NN], parn[NN], Hash[NN], m,
       segtree[NN];
vector<int> G[NN];
void DFS(int ver, int par, int cnt) {
  int in = ti++;
  for (int u : G[ver])
    if (u != par) DFS(u, ver, cnt + 1);
  Hash[ver] = ++m;
  lev[Hash[ver]] = cnt;
  parn[Hash[ver]] = par;
  timein[Hash[ver]] = in;
  timeout[Hash[ver]] = ti++;
}
bool isgra(int ver, int par) {
  return (timein[par] <= timein[ver] && timeout[par] >= timeout[ver]);
}
unordered_map<int, unordered_map<int, int> > memo;
int com_par(int u, int v) {
  if (isgra(v, u)) return u;
  if (isgra(u, v)) return v;
  auto it1 = memo.find(u);
  if (it1 != memo.end()) {
    auto it2 = it1->second.find(v);
    if (it2 != it1->second.end()) return it2->second;
  }
  return memo[u][v] = memo[v][u] = com_par(u, parn[v]);
}
void build(int pos, int st, int en) {
  if (st == en) {
    segtree[pos] = st;
    return;
  }
  build(2 * pos, st, (st + en) / 2);
  build(2 * pos + 1, (st + en) / 2 + 1, en);
  segtree[pos] = com_par(segtree[2 * pos], segtree[2 * pos + 1]);
}
int seg_com_par(int pos, int st, int en, int u, int v) {
  if (en < u || st > v) return -1;
  if (st >= u && en <= v) return segtree[pos];
  int flag1 = seg_com_par(2 * pos, st, (st + en) / 2, u, v);
  int flag2 = seg_com_par(2 * pos + 1, (st + en) / 2 + 1, en, u, v);
  int res;
  if (flag1 != -1 && flag2 != -1)
    res = com_par(flag1, flag2);
  else if (flag1 != -1)
    res = flag1;
  else
    res = flag2;
  return res;
}
int dis(int v, int u) {
  if (u > v) swap(u, v);
  return lev[v] + lev[u] - 2 * lev[seg_com_par(1, 1, m, u, v)];
}
int main() {
  cin >> n;
  int u, v;
  for (int i = (1); i <= (n - 1); ++i) {
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  DFS(1, -1, 0);
  parn[Hash[1]] = 1;
  for (int i = (1); i <= (n); ++i) parn[i] = Hash[parn[i]];
  build(1, 1, m);
  int x, y, a, b, k, q;
  cin >> q;
  while (q--) {
    scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
    x = Hash[x];
    y = Hash[y];
    a = Hash[a];
    b = Hash[b];
    int flag = 0, ab = dis(a, b), axyb = 1 + dis(a, x) + dis(y, b),
        ayxb = 1 + dis(a, y) + dis(x, b);
    if (ab % 2 == k % 2 && ab <= k) flag = 1;
    if (axyb % 2 == k % 2 && axyb <= k) flag = 1;
    if (ayxb % 2 == k % 2 && ayxb <= k) flag = 1;
    printf(flag ? "YES\n" : "NO\n");
  }
  return 0;
}
