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;
using ll = long long;
using ld = long double;
template <typename T>
using V = vector<T>;
template <typename T>
using VV = vector<vector<T>>;
const ll mod = 1000000007;
const ll INF = 1000000099;
const ll LINF = (ll)(1e18 + 99);
const vector<ll> dx = {-1, 1, 0, 0}, dy = {0, 0, -1, 1};
template <typename T, typename U>
inline bool chmin(T& t, const U& u) {
  if (t > u) {
    t = u;
    return 1;
  }
  return 0;
}
template <typename T, typename U>
inline bool chmax(T& t, const U& u) {
  if (t < u) {
    t = u;
    return 1;
  }
  return 0;
}
template <typename T>
inline T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T, typename Y>
inline T mpow(T a, Y n) {
  T res = 1;
  for (; n; n >>= 1) {
    if (n & 1) res = res * a;
    a = a * a;
  }
  return res;
}
template <typename T>
struct LCA {
  int root, lg;
  vector<vector<T>> e;
  vector<T> dps;
  vector<vector<T>> anc;
  LCA(int _root, vector<vector<T>>& _e)
      : root(_root), lg(32 - __builtin_clz(_e.size())), e(_e), dps(_e.size()) {
    int numV = _e.size();
    anc.assign(lg, vector<T>(numV, -1));
    dfs(root, -1, 0);
    for (int i = 0; i < lg - 1; i++) {
      for (int j = 0; j < numV; j++) {
        if (anc[i][j] < 0)
          anc[i + 1][j] = -1;
        else
          anc[i + 1][j] = anc[i][anc[i][j]];
      }
    }
  }
  void dfs(int now, int par, int d) {
    anc[0][now] = par;
    dps[now] = d;
    for (int i = 0; i < (int)e[now].size(); i++) {
      if (e[now][i] != par) dfs(e[now][i], now, d + 1);
    }
  }
  int get_lca(int a, int b) {
    if (dps[a] < dps[b]) swap(a, b);
    int dif = dps[a] - dps[b];
    for (int i = lg - 1; 0 <= i; i--) {
      if (dif < (1 << i)) continue;
      dif -= (1 << i);
      a = anc[i][a];
    }
    if (a == b) return a;
    for (int i = lg - 1; 0 <= i; i--) {
      if (anc[i][a] != anc[i][b]) {
        a = anc[i][a];
        b = anc[i][b];
      }
    }
    return anc[0][a];
  }
};
V<int> dis;
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);
  ll n;
  cin >> n;
  VV<int> e(n, V<int>(0));
  int x, y, a, b, k;
  for (ll i = 0; i < (ll)(n - 1); ++i) {
    cin >> a >> b;
    a--;
    b--;
    e[a].push_back(b);
    e[b].push_back(a);
  }
  dis.assign(n, INF);
  queue<int> q;
  q.push(0);
  q.push(0);
  dis[0] = 0;
  while (q.size()) {
    int now = q.front();
    q.pop();
    int d = q.front();
    q.pop();
    for (ll i = 0; i < (ll)((ll)(e[now]).size()); ++i) {
      if (dis[e[now][i]] == INF) {
        dis[e[now][i]] = d + 1;
        q.push(e[now][i]);
        q.push(d + 1);
      }
    }
  }
  LCA<int> lca(0, e);
  int qq;
  cin >> qq;
  bool can = false;
  ll da, db, loop;
  while (qq--) {
    cin >> x >> y >> a >> b >> k;
    x--;
    y--;
    a--;
    b--;
    can = false;
    int l = lca.get_lca(a, b);
    if ((dis[a] + dis[b] - 2 * dis[l]) % 2 == k % 2 &&
        (dis[a] + dis[b] - 2 * dis[l]) <= k)
      can = true;
    da = dis[a] + dis[x] - 2 * dis[lca.get_lca(a, x)];
    db = dis[y] + dis[b] - 2 * dis[lca.get_lca(y, b)];
    if ((da + db + 1) % 2 == k % 2 && (da + db + 1) <= k) can = true;
    da = dis[a] + dis[y] - 2 * dis[lca.get_lca(a, y)];
    db = dis[x] + dis[b] - 2 * dis[lca.get_lca(x, b)];
    if ((da + db + 1) % 2 == k % 2 && (da + db + 1) <= k) can = true;
    if (can)
      cout << "YES" << '\n';
    else
      cout << "NO" << '\n';
  }
}
