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 nmax = 1e5 + 5;
const int pmax = 17;
vector<int> l[nmax];
int tata[nmax], lca[nmax][30], d[nmax];
void dfs(int nod, int t) {
  lca[nod][0] = t;
  d[nod] = d[t] + 1;
  for (int fiu : l[nod])
    if (fiu != t) dfs(fiu, nod);
}
void build_lca(int n) {
  for (int i = 1; i <= pmax; i++)
    for (int j = 1; j <= n; j++) lca[j][i] = lca[lca[j][i - 1]][i - 1];
}
int find_lca(int x, int y) {
  int l = 0;
  if (d[x] < d[y]) swap(x, y);
  int dd = d[x] - d[y];
  while (dd) {
    int z = dd & (-dd);
    int xe = (int)log2(z);
    x = lca[x][xe];
    dd -= z;
    l += z;
  }
  if (x == y) return l;
  l += 2;
  for (int i = pmax; i >= 0; i--)
    if (lca[x][i] != lca[y][i]) {
      x = lca[x][i];
      y = lca[y][i];
      l += (1 << (i + 1));
    }
  return l;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, u, v, q, x, y, k;
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    l[u].push_back(v);
    l[v].push_back(u);
  }
  dfs(1, 0);
  build_lca(n);
  cin >> q;
  while (q--) {
    int mn = INT_MAX;
    cin >> x >> y >> u >> v >> k;
    int lcadir = find_lca(u, v);
    int lcaxy = find_lca(u, x) + find_lca(y, v) + 1;
    int lcayx = find_lca(u, y) + find_lca(x, v) + 1;
    if ((lcadir & 1) == (k & 1)) mn = min(mn, lcadir);
    if ((lcaxy & 1) == (k & 1)) mn = min(mn, lcaxy);
    if ((lcayx & 1) == (k & 1)) mn = min(mn, lcayx);
    cout << ((mn <= k) ? "YES" : "NO") << "\n";
  }
  return 0;
}
