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 int maxn = 100005;
int n, q;
vector<int> g[maxn];
int d[maxn];
int deda[maxn][20];
int brojac;
int in[maxn];
int out[maxn];
void dfs(int v, int p) {
  deda[v][0] = p;
  for (int i = 1; i < 18; i++) deda[v][i] = deda[deda[v][i - 1]][i - 1];
  in[v] = ++brojac;
  for (auto u : g[v]) {
    if (u != p) {
      d[u] = d[v] + 1;
      dfs(u, v);
    }
  }
  out[v] = brojac;
}
bool predak(int u, int v) { return (in[u] <= in[v] && out[u] >= out[v]); }
int lca(int u, int v) {
  if (predak(u, v)) return u;
  if (predak(u, v)) return v;
  for (int i = 18; i >= 0; i--) {
    if (deda[u][i] && !predak(deda[u][i], v)) u = deda[u][i];
  }
  return deda[u][0];
}
int dist(int u, int v) { return d[u] + d[v] - 2 * d[lca(u, v)]; }
long long calc(int u) {
  for (int i = 18; i >= 0; i--) {
    if (u & (1 << i)) u = deda[u][i];
  }
  return u;
}
bool check(int x, int k) { return (x <= k && (x % 2 == k % 2)); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  cin >> q;
  while (q--) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    int kec = dist(a, b);
    int dva = dist(a, x) + dist(b, y) + 1;
    int tri = dist(b, x) + dist(a, y) + 1;
    if (check(kec, k) || check(dva, k) || check(tri, k))
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
  return 0;
}
