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 = 100005;
const int M = 20;
void Dfs_Find_Center_Gravity(int x, vector<int> *a, int n, int *ChildNum,
                             int *visit, int *ans, int *pos) {
  int temp;
  for (int i = 0; i < a[x].size(); i++)
    if (visit[a[x][i]] == 0) {
      visit[a[x][i]] = 1;
      Dfs_Find_Center_Gravity(a[x][i], a, n, ChildNum, visit, ans, pos);
      temp = max(temp, ChildNum[a[x][i]]);
      ChildNum[x] += ChildNum[a[x][i]] + 1;
    }
  temp = max(temp, n - 1 - ChildNum[x]);
  if (temp < *ans) {
    *ans = temp;
    *pos = x;
  }
  return;
}
int Find_Center_Gravity(vector<int> *a, int n) {
  int ans = n, pos = 0;
  int ChildNum[N], visit[N];
  memset(ChildNum, 0, sizeof(ChildNum));
  memset(visit, 0, sizeof(visit));
  visit[1] = 1;
  Dfs_Find_Center_Gravity(1, a, n, ChildNum, visit, &ans, &pos);
  return pos;
}
void Get_Father(vector<int> *a, int n, int root, int **father, int *depth) {
  int visit[N];
  memset(visit, 0, sizeof(visit));
  visit[root] = 1;
  queue<int> Q;
  Q.push(root);
  while (!Q.empty()) {
    int x = Q.front();
    Q.pop();
    for (int i = 0; i < a[x].size(); i++)
      if (visit[a[x][i]] == 0) {
        visit[a[x][i]] = 1;
        father[a[x][i]][0] = x;
        Q.push(a[x][i]);
        depth[a[x][i]] = depth[x] + 1;
      }
  }
  for (int deep = 0; (1 << deep) < n; deep++)
    for (int i = 1; i <= n; i++)
      if (father[father[i][deep]][deep] != 0)
        father[i][deep + 1] = father[father[i][deep]][deep];
  return;
}
int LCA(int x, int y, int **father, int *depth) {
  if (depth[x] > depth[y]) swap(x, y);
  int delta_depth = depth[y] - depth[x];
  int deep = 0;
  while (delta_depth) {
    if (delta_depth % 2) y = father[y][deep];
    deep++;
    delta_depth /= 2;
  }
  if (x == y) return x;
  deep = (int)(log(depth[x] - 1) / log(2));
  for (int i = deep; i >= 0; i--)
    if (father[x][i] != father[y][i]) {
      x = father[x][i];
      y = father[y][i];
    }
  return father[x][0];
}
int dist(int x, int y, int **father, int *depth) {
  int lca = LCA(x, y, father, depth);
  return depth[x] + depth[y] - 2 * depth[lca];
}
int main() {
  vector<int> *a = new vector<int>[N];
  int *depth = new int[N];
  int **father = new int *[N];
  for (int i = 0; i < N; i++) {
    father[i] = new int[M];
    memset(father[i], 0, M * sizeof(int));
  }
  int n, gra;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  gra = 1;
  int root = gra;
  memset(depth, 0, sizeof(depth));
  depth[root] = 1;
  Get_Father(a, n, root, father, depth);
  int q;
  cin >> q;
  while (q--) {
    int x, y, a, b, K;
    cin >> x >> y >> a >> b >> K;
    int length_path_1 = dist(a, b, father, depth);
    int length_path_2 =
        dist(a, x, father, depth) + 1 + dist(y, b, father, depth);
    int length_path_3 =
        dist(a, y, father, depth) + 1 + dist(x, b, father, depth);
    int length_circle = dist(x, y, father, depth) + 1;
    if (K % 2 == length_path_1 % 2 && K >= length_path_1) {
      cout << "YES" << endl;
      continue;
    }
    if (K % 2 == length_path_2 % 2 && K >= length_path_2) {
      cout << "YES" << endl;
      continue;
    }
    if (K % 2 == length_path_3 % 2 && K >= length_path_3) {
      cout << "YES" << endl;
      continue;
    }
    cout << "NO" << endl;
  }
}
