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 inf = 1 << 30;
const long long INF = 1ll << 60;
const double Inf = 1e20;
const double eps = 1e-9;
void gmax(int &a, int b) { a = (a > b ? a : b); }
void gmin(int &a, int b) { a = (a < b ? a : b); }
const int maxn = 100050;
int n, dep[maxn], fa[maxn][20];
vector<int> adj[maxn];
void dfs(int u) {
  for (auto p : adj[u])
    if (!dep[p]) {
      dep[p] = dep[u] + 1;
      fa[p][0] = u;
      dfs(p);
    }
}
int get_lca(int u, int v) {
  if (dep[u] < dep[v]) swap(u, v);
  int c = (dep[u] - dep[v]);
  for (int j = 19; j >= 0; --j)
    if (c & (1 << j)) u = fa[u][j];
  for (int j = 19; j >= 0; --j)
    if (fa[u][j] != fa[v][j]) u = fa[u][j], v = fa[v][j];
  if (u != v)
    return fa[u][0];
  else
    return u;
}
int calc(int x, int y) { return dep[x] + dep[y] - 2 * dep[get_lca(x, y)]; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dep[1] = 1;
  dfs(1);
  for (int j = 1; j <= 19; ++j)
    for (int i = 1; i <= n; ++i) fa[i][j] = fa[fa[i][j - 1]][j - 1];
  int q;
  scanf("%d", &q);
  while (q--) {
    int x, y, a, b, k;
    scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
    if (calc(a, b) <= k && (calc(a, b) % 2) == (k % 2))
      printf("YES\n");
    else if ((calc(a, x) + 1 + calc(y, b)) <= k &&
             (calc(a, x) + 1 + calc(y, b)) % 2 == (k % 2))
      printf("YES\n");
    else if ((calc(a, y) + 1 + calc(x, b)) <= k &&
             (calc(a, y) + 1 + calc(x, b)) % 2 == (k % 2))
      printf("YES\n");
    else if ((calc(a, x) + calc(x, y) + 1 + calc(x, b)) <= k &&
             (calc(a, x) + calc(x, y) + 1 + calc(x, b)) % 2 == (k % 2))
      printf("YES\n");
    else if ((calc(a, y) + calc(y, x) + 1 + calc(y, b)) <= k &&
             (calc(a, y) + calc(y, x) + 1 + calc(y, b)) % 2 == (k % 2))
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}
