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;
int a, b, k, t, x, y, n, q, i, j, up[N][20], ln, dist[N], tin[N], tout[N];
vector<int> v[N];
void dfs(int u, int pr) {
  tin[u] = ++t;
  up[u][0] = pr;
  for (i = 1; i <= ln; ++i) up[u][i] = up[up[u][i - 1]][i - 1];
  for (auto j : v[u])
    if (pr != j) {
      dist[j] = dist[u] + 1;
      dfs(j, u);
    }
  tout[u] = ++t;
}
int check(int x, int y) { return tin[x] <= tin[y] && tout[x] >= tout[y]; }
int lca(int x, int y) {
  if (check(x, y)) return x;
  if (check(y, x)) return y;
  for (j = ln; j >= 0; --j) {
    if (!check(up[x][j], y)) x = up[x][j];
  }
  return up[x][0];
}
int get(int x, int y) {
  int h = lca(x, y);
  return dist[x] + dist[y] - 2 * dist[h];
}
int main() {
  cin >> n;
  for (i = 1; i < n; ++i) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  while ((1 << ln) <= n) ++ln;
  ++ln;
  dfs(1, 1);
  cin >> q;
  for (i = 1; i <= q; ++i) {
    cin >> x >> y >> a >> b >> k;
    vector<int> ans;
    ans.push_back(get(a, b));
    ans.push_back(get(a, x) + get(b, y) + 1);
    ans.push_back(get(a, y) + get(b, x) + 1);
    int f = 0;
    for (auto j : ans) {
      if (j <= k && (k - j) % 2 == 0) f = 1;
    }
    if (f)
      cout << "YES\n";
    else
      cout << "NO\n";
  }
}
