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;
vector<int> aray[100001];
vector<int> euler;
int dist[100001];
void dfs(int root, int count) {
  dist[root] = count;
  euler.push_back(root);
  for (int i = 0; i < aray[root].size(); i++) {
    int val = aray[root][i];
    if (dist[val] == -1) {
      dfs(val, count + 1);
      euler.push_back(root);
    }
  }
  return;
}
bool valid(int a, int b) {
  if (a > b) return false;
  if (((a - b) % 2) == 0) return true;
  return false;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  int t = n - 1, x, y;
  while (t--) {
    cin >> x >> y;
    aray[x].push_back(y);
    aray[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) dist[i] = -1;
  dfs(1, 0);
  int first[n + 1];
  for (int i = euler.size() - 1; i >= 0; i--) first[euler[i]] = i;
  int m = euler.size();
  int s = sqrt(m);
  int box = ceil((double)m / s);
  int num[box];
  for (int i = 0; i < box; i++) {
    num[i] = 1000000000;
    for (int j = i * s; j < min(m, (i + 1) * s); j++)
      num[i] = min(num[i], dist[euler[j]]);
  }
  int q;
  cin >> q;
  while (q--) {
    int arr[4], k;
    for (int i = 0; i < 4; i++) cin >> arr[i];
    cin >> k;
    int count = 0;
    bool ans = false;
    vector<int> dia;
    for (int i = 0; i < 2; i++) {
      for (int j = 2; j < 4; j++) {
        int fir = first[arr[i]], sec = first[arr[j]];
        int temp = min(fir, sec), tmep = max(fir, sec);
        fir = temp;
        sec = tmep;
        int bfir = fir / s, bsec = sec / s;
        int val = 100000000;
        if (bfir == bsec) {
          for (int i = fir; i <= sec; i++) val = min(val, dist[euler[i]]);
        } else {
          for (int i = bfir + 1; i < bsec; i++) val = min(val, num[i]);
          for (int i = fir; i < bfir * s + s; i++)
            val = min(val, dist[euler[i]]);
          for (int i = s * bsec; i <= sec; i++) val = min(val, dist[euler[i]]);
        }
        int distance = dist[arr[i]] + dist[arr[j]] - 2 * val;
        dia.push_back(distance);
      }
    }
    if (valid(dia[0] + dia[3] + 1, k)) ans = true;
    if (valid(dia[1] + dia[2] + 1, k)) ans = true;
    int fir = first[arr[2]], sec = first[arr[3]];
    int temp = min(fir, sec), tmep = max(fir, sec);
    fir = temp;
    sec = tmep;
    int bfir = fir / s, bsec = sec / s;
    int val = 10000000;
    if (bfir == bsec) {
      for (int i = fir; i <= sec; i++) val = min(val, dist[euler[i]]);
    } else {
      for (int i = bfir + 1; i < bsec; i++) val = min(val, num[i]);
      for (int i = fir; i < bfir * s + s; i++) val = min(val, dist[euler[i]]);
      for (int i = s * bsec; i <= sec; i++) val = min(val, dist[euler[i]]);
    }
    int distance = dist[arr[2]] + dist[arr[3]] - 2 * val;
    if (valid(distance, k)) ans = true;
    if ((distance <= k) && ((distance - k) % 2) == 0) ans = true;
    if (ans)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
  return 0;
}
