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>
#pragma GCC optimize("Ofast")
using namespace std;
const long long int MOD = INT_MAX;
const double PI = acos(-1);
void openfile() {
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
}
const int N = 1e5 + 5;
vector<int> G[N];
int parent[25][N];
int depth[N];
int n;
void dfs(int v, int par, int d) {
  parent[0][v] = par;
  depth[v] = d;
  for (int i = 0; i < ((long long int)G[v].size()); i++) {
    if (G[v][i] != par) dfs(G[v][i], v, d + 1);
  }
}
void init() {
  dfs(1, -1, 0);
  for (int k = 0; k < 20; k++) {
    for (int i = 1; i <= n; i++) {
      if (parent[k][i] < 0)
        parent[k + 1][i] = -1;
      else
        parent[k + 1][i] = parent[k][parent[k][i]];
    }
  }
}
int lca(int u, int v) {
  if (depth[u] > depth[v]) swap(u, v);
  for (int k = 0; k < 20; k++) {
    if ((depth[v] - depth[u]) >> k & 1) v = parent[k][v];
  }
  if (u == v) return u;
  for (int k = 19; k >= 0; k--) {
    if (parent[k][u] != parent[k][v]) {
      u = parent[k][u];
      v = parent[k][v];
    }
  }
  return parent[0][u];
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long int i = 0; i < (n - 1); i++) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  init();
  int q;
  cin >> q;
  for (long long int qq = 0; qq < (q); qq++) {
    int a, b, x, y, d;
    cin >> a >> b >> x >> y >> d;
    int xy_lca = lca(x, y);
    int ax_lca = lca(a, x);
    int ay_lca = lca(a, y);
    int bx_lca = lca(b, x);
    int by_lca = lca(b, y);
    int xy_length = depth[x] + depth[y] - depth[xy_lca] * 2;
    int ax_length = depth[a] + depth[x] - depth[ax_lca] * 2;
    int ay_length = depth[a] + depth[y] - depth[ay_lca] * 2;
    int bx_length = depth[b] + depth[x] - depth[bx_lca] * 2;
    int by_length = depth[b] + depth[y] - depth[by_lca] * 2;
    bool ok = 0;
    if (xy_length <= d && (d - xy_length) % 2 == 0) ok = 1;
    if (ax_length + by_length + 1 <= d &&
        (d - (ax_length + by_length + 1)) % 2 == 0)
      ok = 1;
    if (ay_length + bx_length + 1 <= d &&
        (d - (ay_length + bx_length + 1)) % 2 == 0)
      ok = 1;
    if (ok)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
}
