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 long long int maxn = 1000005;
long long int tp[1000005];
int tin[maxn];
int tout[maxn];
int curt = 0;
int const maxl = 19;
int par[maxn][maxl];
int deep[maxn];
vector<long long int> ve[1000005];
void dfs(const int v, const int p = 1) {
  tin[v] = curt++;
  par[v][0] = p;
  deep[v] = deep[p] + 1;
  for (int i = 1; i < maxl; ++i) par[v][i] = par[par[v][i - 1]][i - 1];
  for (int ne : ve[v])
    if (ne != p) dfs(ne, v);
  tout[v] = curt++;
}
inline bool upper(const int x, const int y) {
  return tin[x] <= tin[y] && tout[y] <= tout[x];
}
inline int lca(int x, int y) {
  if (upper(x, y)) return x;
  if (upper(y, x)) return y;
  for (int i = maxl - 1; i >= 0; --i)
    if (!upper(par[x][i], y)) x = par[x][i];
  return par[x][0];
}
inline int dist(const int x, const int y) {
  return deep[x] + deep[y] - 2 * deep[lca(x, y)];
}
bool vis[1000005];
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n;
  cin >> n;
  for (long long int i = 1; i <= n - 1; i++) {
    long long int u, v;
    cin >> u >> v;
    ve[u].push_back(v);
    ve[v].push_back(u);
  }
  dfs(1);
  long long int qq;
  cin >> qq;
  while (qq--) {
    long long int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    long long int dis = dist(a, b);
    if (dis % 2 == k % 2) {
      if (k >= dis) {
        cout << "YES"
             << " " << endl;
        ;
        continue;
      }
    }
    long long int minn = dist(a, x) + dist(y, b) + 1;
    if (minn % 2 == k % 2) {
      if (k >= minn) {
        cout << "YES"
             << " " << endl;
        ;
        continue;
      }
    }
    minn = dist(b, x) + dist(y, a) + 1;
    if (minn % 2 == k % 2) {
      if (k >= minn) {
        cout << "YES"
             << " " << endl;
        ;
        continue;
      }
    }
    cout << "NO"
         << " " << endl;
    ;
  }
  return 0;
}
