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 maxl = 1e5 + 10;
int n, q, cnt;
int ehead[maxl], dep[maxl];
int f[21][maxl];
struct ed {
  int to, nxt;
} e[maxl << 1];
inline void add(int u, int v) {
  e[++cnt].to = v;
  e[cnt].nxt = ehead[u];
  ehead[u] = cnt;
}
inline void dfs(int u, int fa) {
  f[0][u] = fa;
  int v;
  for (int i = ehead[u]; i; i = e[i].nxt) {
    v = e[i].to;
    if (v == fa) continue;
    dep[v] = dep[u] + 1;
    dfs(v, u);
  }
}
inline void prework() {
  scanf("%d", &n);
  int u, v;
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  dfs(1, 0);
  for (int i = 1; i <= 20; i++)
    for (int j = 1; j <= n; j++) f[i][j] = f[i - 1][f[i - 1][j]];
}
inline int lca(int u, int v) {
  if (dep[v] > dep[u]) swap(u, v);
  for (int i = 0; i <= 20; i++)
    if ((1 << i) & (dep[u] - dep[v])) u = f[i][u];
  if (u == v) return u;
  for (int i = 20; i >= 0; i--)
    if (f[i][u] != f[i][v]) u = f[i][u], v = f[i][v];
  return f[0][u];
}
inline int dist(int u, int v) { return dep[u] + dep[v] - 2 * dep[lca(u, v)]; }
inline void mainwork() {
  scanf("%d", &q);
  int a, b, x, y, k, len;
  bool flag;
  a = lca(2, 3);
  for (int i = 1; i <= q; i++) {
    scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
    flag = false;
    len = dist(a, b);
    if ((len & 1) == (k & 1) && len <= k) flag = true;
    len = dist(a, x) + dist(b, y) + 1;
    if ((len & 1) == (k & 1) && len <= k) flag = true;
    len = dist(a, y) + dist(b, x) + 1;
    if ((len & 1) == (k & 1) && len <= k) flag = true;
    if (flag)
      puts("YES");
    else
      puts("NO");
  }
}
int main() {
  prework();
  mainwork();
  return 0;
}
