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>
const long long int MOD = 1e9 + 7;
using namespace std;
std::vector<int> adj[200005];
int dp[200005][20], dis[200005] = {0}, tin[200005], tout[200005], tim = 0;
void dfs(int u, int par) {
  tin[u] = ++tim;
  for (int v : adj[u]) {
    if (v == par) continue;
    dp[v][0] = u;
    dis[v] = dis[u] + 1;
    dfs(v, u);
  }
  tout[u] = ++tim;
}
bool is_ancestor(int u, int v) {
  if (tin[u] <= tin[v] && tout[u] >= tout[v]) return true;
  return false;
}
int LCA(int u, int v) {
  if (is_ancestor(u, v)) return u;
  if (is_ancestor(v, u)) return v;
  for (int i = 19; i >= 0; i--) {
    if (!is_ancestor(dp[u][i], v)) u = dp[u][i];
  }
  return dp[u][0];
}
int cal_dis(int a, int b) {
  int lca = LCA(a, b);
  return dis[a] + dis[b] - 2 * dis[lca];
}
bool solve(long long int x, long long int y, long long int a, long long int b,
           long long int k) {
  int dist = cal_dis(a, b);
  if (k >= dist && (k - dist) % 2 == 0) return true;
  dist = cal_dis(a, x) + 1 + cal_dis(y, b);
  if (k >= dist && (k - dist) % 2 == 0) return true;
  dist = cal_dis(a, y) + 1 + cal_dis(x, b);
  if (k >= dist && (k - dist) % 2 == 0) return true;
  return false;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n;
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, 1);
  dp[1][0] = 1;
  for (int i = 1; i < 20; i++) {
    for (int j = 1; j <= n; j++) dp[j][i] = dp[dp[j][i - 1]][i - 1];
  }
  int q;
  cin >> q;
  while (q--) {
    long long int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    if (solve(x, y, a, b, k))
      cout << "YES" << '\n';
    else
      cout << "NO" << '\n';
  }
  return 0;
}
