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 MOD = 1e9 + 7;
const long pp = 1e6;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
long long n, m, k;
int dad[18][pp], h[pp];
vector<int> V[pp];
string S;
map<long long, long long> M;
set<long long> Set;
void dfs(int u) {
  for (auto v : V[u])
    if (v != dad[0][u]) {
      dad[0][v] = u;
      h[v] = h[u] + 1;
      dfs(v);
    }
}
int lca(int u, int v) {
  if (h[u] < h[v]) swap(u, v);
  for (int i = 17; i >= 0; --i)
    if (((h[u] - h[v]) >> i) & 1) u = dad[i][u];
  if (u == v) return u;
  for (int i = 17; i >= 0; --i)
    if (dad[i][u] != dad[i][v]) u = dad[i][u], v = dad[i][v];
  return dad[0][u];
}
int get(int u, int v) { return (h[u] + h[v] - 2 * h[lca(u, v)]); }
bool ok(int x, int y) { return y >= x && x % 2 == y % 2; }
void dowork() {
  cin >> n;
  for (int i = 1; i <= n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    V[u].push_back(v);
    V[v].push_back(u);
  }
  dfs(1);
  for (int i = 1; i <= 17; ++i)
    for (int j = 1; j <= n; ++j) dad[i][j] = dad[i - 1][dad[i - 1][j]];
  cin >> m;
  for (int i = 1; i <= m; ++i) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    int path1 = get(a, b);
    int path2 = get(a, x) + get(b, y) + 1;
    int path3 = get(a, y) + get(b, x) + 1;
    if (ok(path1, k) || ok(path2, k) || ok(path3, k)) {
      cout << "YES";
    } else
      cout << "NO";
    cout << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  dowork();
  return 0;
}
