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 = 1e5 + 41;
const int mod = 1e9 + 7;
const int inf = 1e9 + 90;
const long long INF = 1e18;
vector<int> g[N];
int tin[N], tout[N], up[N][20], d[N], timer, a, b;
void dfs(int v = 1, int p = 1) {
  tin[v] = ++timer;
  up[v][0] = p;
  for (int i = 1; i <= 18; ++i) {
    up[v][i] = up[up[v][i - 1]][i - 1];
  }
  for (int to : g[v])
    if (to != p) {
      d[to] = d[v] + 1;
      dfs(to, v);
    }
  tout[v] = ++timer;
}
bool upper(int a, int b) { return tin[a] <= tin[b] && tout[a] >= tout[b]; }
int lca(int a, int b) {
  if (upper(a, b)) return a;
  if (upper(b, a)) return b;
  for (int i = 18; i >= 0; --i) {
    if (!upper(up[a][i], b)) {
      a = up[a][i];
    }
  }
  return up[a][0];
}
int dist(int a, int b) { return d[a] + d[b] - 2 * d[lca(a, b)]; }
bool ok(int dist1, int dist2) {
  return dist1 >= dist2 && ((dist1 & 1) == (dist2 & 1));
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  ;
  int n;
  cin >> n;
  for (int i = 1; i < n; ++i) {
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs();
  int q;
  cin >> q;
  while (q--) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    bool res = 0;
    res |= ok(k, dist(a, b));
    res |= ok(k, dist(a, x) + 1 + dist(y, b));
    res |= ok(k, dist(a, y) + 1 + dist(x, b));
    res |= ok(k, dist(x, y) + 1 + dist(a, x) + dist(x, b));
    res |= ok(k, dist(x, y) + 1 + dist(a, y) + dist(y, b));
    cout << (res ? "YES\n" : "NO\n");
  }
}
