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 long long maxn = 1e5 + 5, mod = 1e9 + 7, inf = 0x3f3f3f3f;
long long n, m, Q, K, T;
long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * f;
}
long long e = 1, to[maxn << 1], hed[maxn], nxt[maxn << 1];
void add(long long x, long long y) {
  to[++e] = y;
  nxt[e] = hed[x];
  hed[x] = e;
  to[++e] = x;
  nxt[e] = hed[y];
  hed[y] = e;
}
long long fa[maxn], dep[maxn], siz[maxn], son[maxn], top[maxn];
void dfs1(long long x, long long fu, long long deep) {
  siz[x] = 1, fa[x] = fu;
  dep[x] = deep;
  for (long long i = hed[x]; i; i = nxt[i]) {
    if (to[i] == fu) continue;
    dfs1(to[i], x, deep + 1);
    siz[x] += siz[to[i]];
    if (siz[to[i]] > siz[son[x]]) son[x] = to[i];
  }
  return;
}
void dfs2(long long x, long long topx) {
  top[x] = topx;
  if (!son[x]) return;
  dfs2(son[x], topx);
  for (long long i = hed[x]; i; i = nxt[i]) {
    if (to[i] == fa[x] || to[i] == son[x]) continue;
    dfs2(to[i], to[i]);
  }
  return;
}
long long lca(long long x, long long y) {
  while (top[x] != top[y]) {
    if (dep[top[x]] < dep[top[y]]) swap(x, y);
    x = fa[top[x]];
  }
  if (dep[x] > dep[y]) swap(x, y);
  return x;
}
long long dis(long long x, long long y) {
  return dep[x] + dep[y] - dep[lca(x, y)] * 2;
}
bool check(long long x, long long y, long long a, long long b, long long k) {
  long long goal = (k & 1);
  long long k1 = ((dis(a, b))), k2 = ((dis(a, x) + dis(b, y) + 1));
  long long k3 = ((dis(a, y) + dis(b, x) + 1));
  if (k1 <= k && k1 % 2 == goal) return true;
  if (k2 <= k && k2 % 2 == goal) return true;
  if (k3 <= k && k3 % 2 == goal) return true;
  return false;
}
signed main() {
  n = read();
  for (long long i = (1); i <= (n - 1); ++i) add(read(), read());
  dfs1(1, 0, 1);
  dfs2(1, 1);
  Q = read();
  while (Q--) {
    long long x = read(), y = read(), a = read(), b = read(), k = read();
    puts((check(x, y, a, b, k)) ? "YES" : "NO");
  }
  return 0;
}
