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 + 5;
const int inf = 0x3f3f3f3f;
struct node {
  int to, next;
} edge[maxn << 2];
int head[maxn], dep[maxn], fa[maxn][20], cnt, n;
void add(int a, int b) {
  edge[++cnt].to = b;
  edge[cnt].next = head[a];
  head[a] = cnt;
}
void dfs(int now, int f) {
  dep[now] = dep[f] + 1;
  fa[now][0] = f;
  for (int i = head[now]; i != -1; i = edge[i].next) {
    int to = edge[i].to;
    if (to == f) continue;
    dfs(to, now);
  }
}
int lca(int a, int b) {
  if (dep[a] < dep[b]) swap(a, b);
  for (int i = 17; i >= 0; --i) {
    if (dep[fa[a][i]] >= dep[b]) {
      a = fa[a][i];
    }
  }
  for (int i = 17; i >= 0; --i) {
    if (fa[a][i] != fa[b][i]) {
      a = fa[a][i];
      b = fa[b][i];
    }
  }
  if (a != b) return fa[a][0];
  return a;
}
int dis(int a, int b) { return dep[a] + dep[b] - 2 * dep[lca(a, b)]; }
int main() {
  int m, a, b, x, y, k;
  memset(head, -1, sizeof(head));
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    scanf("%d %d", &a, &b);
    add(a, b), add(b, a);
  }
  dfs(1, 0);
  for (int i = 1; i <= 17; ++i) {
    for (int j = 1; j <= n; ++j) {
      fa[j][i] = fa[fa[j][i - 1]][i - 1];
    }
  }
  scanf("%d", &m);
  while (m--) {
    scanf("%d %d %d %d %d", &x, &y, &a, &b, &k);
    int len = dis(a, b), ans = 0;
    if (len <= k && ((len & 1) == (k & 1))) {
      ans = 1;
    } else {
      len = dis(a, x) + dis(y, b) + 1;
      if (len <= k && ((len & 1) == (k & 1))) {
        ans = 1;
      }
      len = dis(a, y) + dis(x, b) + 1;
      if (len <= k && ((len & 1) == (k & 1))) {
        ans = 1;
      }
    }
    printf("%s\n", ans ? "YES" : "NO");
  }
  return 0;
}
