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 maxn = 10e5 + 5;
int n;
vector<int> G[maxn];
int f[18][maxn];
int d[maxn];
void dfs(int x) {
  for (int i = 0; i < G[x].size(); i++) {
    int v = G[x][i];
    if (v == f[0][x]) continue;
    f[0][v] = x;
    d[v] = d[x] + 1;
    dfs(v);
  }
}
int lca(int x, int y) {
  if (d[x] < d[y]) swap(x, y);
  for (int i = 17; i >= 0; i--) {
    if (d[x] - (1 << i) >= d[y]) x = f[i][x];
  }
  if (x == y) return x;
  for (int i = 17; i >= 0; i--) {
    if (f[i][x] != f[i][y]) {
      x = f[i][x];
      y = f[i][y];
    }
  }
  return f[0][x];
}
int dis(int x, int y) { return d[x] + d[y] - 2 * d[lca(x, y)]; }
bool check(int x, int y) { return y >= x && ((x % 2) == (y % 2)); }
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfs(1);
  for (int i = 1; i <= 17; i++) {
    for (int j = 1; j <= n; j++) {
      f[i][j] = f[i - 1][f[i - 1][j]];
    }
  }
  int t;
  cin >> t;
  while (t--) {
    int x1, y1, a1, b1, dist1;
    cin >> x1 >> y1 >> a1 >> b1 >> dist1;
    if (check(dis(a1, b1), dist1) ||
        check(dis(a1, x1) + dis(y1, b1) + 1, dist1) ||
        check(dis(a1, y1) + dis(x1, b1) + 1, dist1))
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  }
}
