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 = 1e5 + 10;
int n, q, f[maxn][25], h[maxn];
struct node {
  int nxt, to;
} v[maxn << 1];
int head[maxn << 1], cnt = 1, lg[500001];
void add(int from, int to) {
  v[cnt].to = to;
  v[cnt].nxt = head[from];
  head[from] = cnt++;
}
int lca(int x, int y) {
  if (h[x] < h[y]) swap(x, y);
  while (h[x] > h[y]) x = f[x][lg[h[x] - h[y]] - 1];
  if (x == y) return x;
  for (int k = lg[h[x]] - 1; k >= 0; --k)
    if (f[x][k] != f[y][k]) x = f[x][k], y = f[y][k];
  return f[x][0];
}
int solve(int x, int y) { return h[x] + h[y] - h[lca(x, y)] * 2; }
void dfs(int now, int fath) {
  f[now][0] = fath;
  h[now] = h[fath] + 1;
  for (int i = 1; i <= lg[h[now]]; ++i) f[now][i] = f[f[now][i - 1]][i - 1];
  for (int i = head[now]; i; i = v[i].nxt)
    if (v[i].to != fath) dfs(v[i].to, now);
}
bool check(int x, int y) {
  if (y >= x && ((x % 2) == (y % 2))) return 1;
  return 0;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) lg[i] = lg[i - 1] + (1 << lg[i - 1] == i);
  for (int i = 1, u, v; i < n; i++) {
    cin >> u >> v;
    add(u, v);
    add(v, u);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= 20; j++) f[i][j] = f[f[i][j - 1]][j - 1];
  cin >> q;
  for (int i = 0; i < q; i++) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    int X = solve(a, b), Y = solve(a, x) + solve(y, b) + 1,
        Z = solve(a, y) + solve(x, b) + 1;
    if (check(X, k) || check(Y, k) || check(Z, k))
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
