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;
long long n, Q;
vector<long long> G[100005];
int Prev[100005][17];
int depth[100005];
void predfs(int v, int p, int d) {
  depth[v] = d;
  Prev[v][0] = p;
  for (int i = 0; i < G[v].size(); i++) {
    if (G[v][i] == p) continue;
    predfs(G[v][i], v, d + 1);
  }
}
int getLCA(int u, int v) {
  int x = u, y = v;
  if (depth[y] > depth[x]) swap(x, y);
  for (int i = 16; i >= 0; i--) {
    if (depth[x] - (1 << i) >= depth[y]) x = Prev[x][i];
  }
  if (x == y) return x;
  for (int i = 16; i >= 0; i--) {
    if (Prev[x][i] != Prev[y][i]) {
      x = Prev[x][i];
      y = Prev[y][i];
    }
  }
  x = Prev[x][0];
  return x;
}
long long getDist(long long u, long long v) {
  long long lca = getLCA(u, v);
  return depth[u] + depth[v] - 2 * depth[lca];
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  long long u, v;
  for (int i = 1; i <= n - 1; i++) {
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  predfs(1, 0, 0);
  for (int i = 1; i < 17; i++) {
    for (int j = 1; j <= n; j++) {
      Prev[j][i] = Prev[Prev[j][i - 1]][i - 1];
    }
  }
  cin >> Q;
  long long x, y, a, b, k;
  for (int q = 0; q < Q; q++) {
    cin >> x >> y >> a >> b >> k;
    long long d = getDist(a, b);
    if (d <= k && d % 2 == k % 2) {
      cout << "YES"
           << "\n";
      continue;
    }
    d = getDist(x, a) + getDist(y, b) + 1;
    if (d <= k && d % 2 == k % 2) {
      cout << "YES"
           << "\n";
      continue;
    }
    d = getDist(x, b) + getDist(y, a) + 1;
    if (d <= k && d % 2 == k % 2) {
      cout << "YES"
           << "\n";
      continue;
    }
    cout << "NO"
         << "\n";
  }
  flush(cout);
  return 0;
}
