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 N = 300010;
int h[N], nxt[N << 1], to[N << 1], sz[N], fa[N], dep[N], bl[N], cnt;
void init() {
  memset(h, 0, sizeof h);
  cnt = 0;
}
void add(int x, int y) {
  to[++cnt] = y;
  nxt[cnt] = h[x];
  h[x] = cnt;
}
void dfs1(int x) {
  sz[x] = 1;
  for (int i = h[x]; i; i = nxt[i])
    if (to[i] != fa[x]) {
      fa[to[i]] = x;
      dep[to[i]] = dep[x] + 1;
      dfs1(to[i]);
      sz[x] += sz[to[i]];
    }
}
void dfs2(int x, int c) {
  int k = 0;
  bl[x] = c;
  for (int i = h[x]; i; i = nxt[i])
    if (to[i] != fa[x] && sz[to[i]] > sz[k]) k = to[i];
  if (!k) return;
  dfs2(k, c);
  for (int i = h[x]; i; i = nxt[i])
    if (to[i] != fa[x] && to[i] != k) dfs2(to[i], to[i]);
}
int lca(int x, int y) {
  while (bl[x] != bl[y]) {
    if (dep[bl[x]] < dep[bl[y]]) swap(x, y);
    x = fa[bl[x]];
  }
  if (dep[x] < dep[y]) return x;
  return y;
}
int dist(int x, int y) { return dep[x] + dep[y] - 2 * dep[lca(x, y)]; }
int n, m;
int main() {
  scanf("%d", &n);
  for (int x, y, i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    add(x, y), add(y, x);
  }
  dfs1(1);
  dfs2(1, 1);
  scanf("%d", &m);
  while (m--) {
    int a, b, k, x, y;
    scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
    int l = dist(a, b);
    int f = min(dist(a, x) + dist(b, y), dist(a, y) + dist(b, x)) + 1;
    if ((k >= l && (k - l) % 2 == 0) || (k >= f && (k - f) % 2 == 0))
      puts("YES");
    else
      puts("NO");
  }
}
