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;
long long MOD = (1e9) + 7;
const long long N = 1e5 + 5;
const long long level = 25;
long long par[N][level], depth[N];
vector<long long> a[N];
void dfs(long long c, long long p) {
  depth[c] = depth[p] + 1;
  par[c][0] = p;
  for (long long i = 1; i < level; i++) par[c][i] = par[par[c][i - 1]][i - 1];
  for (auto ne : a[c]) {
    if (ne == p) continue;
    dfs(ne, c);
  }
}
long long lca(long long u, long long v) {
  if (depth[v] < depth[u]) swap(u, v);
  long long diff = depth[v] - depth[u];
  for (long long i = 0; i < level; i++)
    if ((diff >> i) & 1) v = par[v][i];
  if (u == v) return u;
  for (long long i = level - 1; i >= 0; i--)
    if (par[u][i] != par[v][i]) {
      u = par[u][i];
      v = par[v][i];
    }
  return par[u][0];
}
long long dist(long long u, long long v) {
  if (u == v) return 0;
  long long res = 0;
  if (depth[u] > depth[v]) swap(u, v);
  long long l = lca(u, v);
  return depth[u] + depth[v] - (2 * depth[l]);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long tc = 1;
  for (long long ti = 1; ti <= tc; ti++) {
    long long n;
    cin >> n;
    for (long long i = 1; i < n; i++) {
      long long x, y;
      cin >> x >> y;
      a[x].push_back(y);
      a[y].push_back(x);
    }
    dfs(1, 0);
    long long q;
    cin >> q;
    while (q--) {
      long long x, y, a, b, k;
      cin >> x >> y >> a >> b >> k;
      long long chk = dist(a, b);
      if (chk <= k && !((k - chk) % 2)) {
        cout << "YES" << endl;
        continue;
      }
      chk = dist(a, x) + dist(y, b) + 1;
      if (chk <= k && !((k - chk) % 2)) {
        cout << "YES" << endl;
        continue;
      }
      chk = dist(a, y) + dist(x, b) + 1;
      if (chk <= k && !((k - chk) % 2)) {
        cout << "YES" << endl;
        continue;
      }
      cout << "NO" << endl;
    }
  }
  return 0;
}
