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, x, y, q, a, b, k, in[100009], out[100009], parent[100009][19], timer,
    H[100009];
vector<int> v[100009];
void go(int x) {
  timer++;
  in[x] = timer;
  for (int i = 2; i <= 17; i++) {
    parent[x][i] = parent[parent[x][i - 1]][i - 1];
  }
  for (int i = 0; i < v[x].size(); i++) {
    if (v[x][i] != parent[x][1]) {
      parent[v[x][i]][1] = x;
      go(v[x][i]);
    }
  }
  timer++;
  out[x] = timer;
}
bool check(int a, int b) { return (in[a] <= in[b] && out[b] <= out[a]); }
int lca(int a, int b) {
  if (check(a, b)) return a;
  if (check(b, a)) return b;
  for (int i = 17; i >= 1; i--) {
    if (!check(parent[a][i], b)) a = parent[a][i];
  }
  return parent[a][1];
}
void High(int x, int h) {
  H[x] = h;
  for (int i = 0; i < v[x].size(); i++) {
    if (H[v[x][i]] == 0) High(v[x][i], h + 1);
  }
}
bool find(int a, int b, int k) {
  int p = lca(a, b);
  int depth = H[a] + H[b] - 2 * H[p];
  if (k % 2 == depth % 2 && k >= depth)
    return true;
  else
    return false;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  in[0] = 0;
  out[0] = 3 * n;
  go(1);
  High(1, 1);
  cin >> q;
  while (q--) {
    cin >> x >> y >> a >> b >> k;
    if (find(a, b, k)) {
      cout << "YES" << endl;
      continue;
    }
    int p = lca(a, x);
    int depth = H[a] + H[x] - 2 * H[p];
    depth++;
    if (depth <= k) {
      if (find(y, b, k - depth)) {
        cout << "YES" << endl;
        continue;
      }
    }
    p = lca(a, y);
    depth = H[a] + H[y] - 2 * H[p];
    depth++;
    if (depth <= k) {
      if (find(x, b, k - depth)) {
        cout << "YES" << endl;
        continue;
      }
    }
    cout << "NO" << endl;
  }
  return 0;
}
