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 = 1000 * 100 + 5;
const long long inf = 9223372036854775807;
int n, q, mark[maxn], h[maxn], par[maxn][50];
vector<int> adj[maxn];
void dfs(int v) {
  mark[v] = 1;
  for (auto u : adj[v]) {
    if (mark[u] == 0) {
      mark[u] = 1;
      h[u] = h[v] + 1;
      par[u][0] = v;
      for (int i = 1; i <= 31; i++) {
        par[u][i] = par[par[u][i - 1]][i - 1];
      }
      dfs(u);
    }
  }
}
int lca(int u, int v) {
  if (h[v] < h[u]) {
    swap(u, v);
  }
  for (int i = 31; i >= 0; i--) {
    if (par[v][i] != 0 && h[par[v][i]] >= h[u]) {
      v = par[v][i];
    }
  }
  if (v == u) return v;
  for (int i = 31; i >= 0; i--) {
    if (par[v][i] != par[u][i]) {
      v = par[v][i], u = par[u][i];
    }
  }
  return par[v][0];
}
int lp(int u, int v) {
  int lc = lca(u, v), ans = 0;
  for (int i = 31; i >= 0; i--) {
    if (par[u][i] != 0 && h[par[u][i]] >= h[lc]) {
      ans += pow(2, i);
      u = par[u][i];
    }
    if (par[v][i] != 0 && h[par[v][i]] >= h[lc]) {
      ans += pow(2, i);
      v = par[v][i];
    }
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1);
  cin >> q;
  for (int i = 1; i <= q; i++) {
    int u, v, a, b, k;
    cin >> u >> v >> a >> b >> k;
    int x = lp(a, b), y = lp(a, u) + 1 + lp(v, b), z = lp(a, v) + 1 + lp(u, b);
    if (x <= k && x % 2 == k % 2) {
      cout << "YES";
    } else if (y <= k && y % 2 == k % 2) {
      cout << "YES";
    } else if (z <= k && z % 2 == k % 2) {
      cout << "YES";
    } else {
      cout << "NO";
    }
    cout << endl;
  }
}
