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>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {1, -1, 1, 0, -1, 0, 1, -1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
vector<long long> adj[500005];
long long lev[500005], tr[500005][20];
void dfs(long long u, long long par) {
  long long i, j;
  for (i = 0; i < adj[u].size(); i++) {
    long long p = adj[u][i];
    if (p == par) continue;
    lev[p] = lev[u] + 1;
    tr[p][0] = u;
    dfs(p, u);
  }
}
void build(long long n) {
  memset(tr, -1, sizeof(tr));
  dfs(1, -1);
  long long i, j;
  for (j = 1; (1 << j) <= n; j++) {
    for (i = 1; i <= n; i++) {
      if (tr[i][j - 1] != -1) tr[i][j] = tr[tr[i][j - 1]][j - 1];
    }
  }
}
long long query(long long p, long long q) {
  if (lev[q] > lev[p]) swap(p, q);
  long long i, j;
  for (i = 16; i >= 0; i--) {
    if (lev[p] - (1 << i) >= lev[q]) {
      p = tr[p][i];
    }
  }
  if (p == q) return p;
  for (i = 16; i >= 0; i--) {
    if (tr[p][i] != tr[q][i] && tr[p][i] != -1 && tr[q][i] != -1) {
      p = tr[p][i];
      q = tr[q][i];
    }
  }
  return tr[p][0];
}
long long dis(long long x, long long y) {
  long long p = query(x, y);
  return lev[x] + lev[y] - 2 * lev[p];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tc, n, i, j, m, q;
  while (cin >> n) {
    for (i = 1; i < n; i++) {
      long long u, v;
      cin >> u >> v;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    build(n);
    cin >> q;
    while (q--) {
      long long x, y, a, b, k;
      cin >> x >> y >> a >> b >> k;
      long long d = dis(a, b);
      if (d <= k && d % 2 == k % 2) {
        cout << "YES" << endl;
        continue;
      }
      long long p1 = dis(x, a) + dis(y, b) + 1;
      long long p2 = dis(x, b) + dis(y, a) + 1;
      if (p1 <= k && p1 % 2 == k % 2) {
        cout << "YES" << endl;
        continue;
      }
      if (p2 <= k && p2 % 2 == k % 2) {
        cout << "YES" << endl;
        continue;
      }
      cout << "NO" << endl;
    }
  }
  return 0;
}
