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;
int n, u, v;
vector<int> e[100005];
int p[100005][21], d[100005], q, x, y, a, b, k;
void dfs(int x, int fa) {
  d[x] = d[fa] + 1;
  p[x][0] = fa;
  for (int i = 1; i <= 20; ++i) {
    p[x][i] = p[p[x][i - 1]][i - 1];
  }
  for (int i = 0; i < e[x].size(); ++i) {
    int u = e[x][i];
    if (u == fa) continue;
    dfs(u, x);
  }
}
int LCA(int u, int v) {
  if (d[u] < d[v]) swap(u, v);
  for (int i = 20; i >= 0; --i) {
    if (d[u] - d[v] >= (1 << i)) u = p[u][i];
  }
  if (u == v) return u;
  for (int i = 20; i >= 0; --i) {
    if (p[u][i] != p[v][i]) {
      u = p[u][i];
      v = p[v][i];
    }
  }
  return p[u][0];
}
int dis(int u, int v) {
  int top = LCA(u, v);
  return d[u] + d[v] - 2 * d[top];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    e[u].push_back(v);
    e[v].push_back(u);
  }
  dfs(1, 1);
  scanf("%d", &q);
  while (q--) {
    scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
    int d1 = dis(a, b);
    int flag = 0, circle = dis(x, y) + 1;
    if (k >= d1 && ((k - d1) & 1) == 0) flag = 1;
    int d2 = dis(a, x) + dis(b, y) + 1;
    if (k >= d2 && ((k - d2) & 1) == 0) flag = 1;
    int d3 = dis(a, y) + dis(b, x) + 1;
    if (k >= d3 && ((k - d3) & 1) == 0) flag = 1;
    if (k >= d2 + circle) {
      int tmp = k - d2 - circle;
      if ((tmp & 1) == 0) flag = 1;
    }
    if (k >= d3 + circle) {
      int tmp = k - d3 - circle;
      if ((tmp & 1) == 0) flag = 1;
    }
    puts(flag ? "YES" : "NO");
  }
}
