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 double PI = acos(-1.0);
const double EPS = 1e-9;
const long long MOD = 1e9 + 7;
const long long INF = 0x3f3f3f3f;
const long long MAXN = 1e5 + 7;
const long long LOG = 20;
int n, q, h[MAXN], f[MAXN][LOG];
vector<int> g[MAXN];
void dfs(int u, int fa, int height) {
  h[u] = height;
  f[u][0] = fa;
  for (int i = 1; i < LOG; ++i) f[u][i] = f[f[u][i - 1]][i - 1];
  for (auto v : g[u]) {
    if (v == fa) continue;
    dfs(v, u, height + 1);
  }
}
int proc(int a, int b) {
  int x = h[a], y = h[b];
  if (x < y) swap(a, b);
  for (int i = LOG - 1; i >= 0; --i) {
    if (h[f[a][i]] >= h[b]) a = f[a][i];
  }
  if (h[a] > h[b]) a = f[a][0];
  for (int i = LOG - 1; i >= 0; --i) {
    if (f[a][i] != f[b][i]) a = f[a][i], b = f[b][i];
  }
  if (a != b) a = f[a][0];
  int z = x - h[a] + y - h[a];
  return z;
}
void solve() {
  cin >> n;
  for (int u, v, i = 1; i < n; ++i) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0, 1);
  int x, y, a, b, k;
  cin >> q;
  while (q--) {
    cin >> x >> y >> a >> b >> k;
    vector<int> path;
    path.push_back(proc(a, b));
    path.push_back(proc(a, x) + 1 + proc(y, b));
    path.push_back(proc(a, y) + 1 + proc(x, b));
    int ans = 0;
    for (auto v : path) ans |= (v <= k && (v & 1) == (k & 1));
    cout << (ans ? "YES" : "NO") << endl;
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int i, Case = 1;
  while (Case--) solve();
  return 0;
}
