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 int N = 100010;
const int K = 17;
int n;
vector<int> adj[N];
int depth[N];
int P[N][K];
void dfs(int u, int p, int d) {
  depth[u] = d;
  P[u][0] = p;
  for (int v : adj[u]) {
    if (v == p) continue;
    dfs(v, u, d + 1);
  }
}
int LCA(int a, int b) {
  if (depth[a] < depth[b]) swap(a, b);
  for (int j = K - 1; j >= 0; --j) {
    if (depth[a] - (1 << j) >= depth[b]) {
      a = P[a][j];
    }
  }
  if (a == b) return a;
  for (int j = K - 1; j >= 0; --j) {
    if (P[a][j] != -1 && P[a][j] != P[b][j]) {
      a = P[a][j], b = P[b][j];
    }
  }
  return P[a][0];
}
inline int dist(int a, int b) {
  return depth[a] + depth[b] - 2 * depth[LCA(a, b)];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, -1, 0);
  for (int j = 1; j < K; ++j) {
    for (int i = 1; i <= n; ++i) {
      if (P[i][j - 1] == -1) {
        P[i][j] = -1;
      } else {
        P[i][j] = P[P[i][j - 1]][j - 1];
      }
    }
  }
  int q;
  cin >> q;
  while (q--) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    int aux;
    int cyclen = 1 + dist(x, y);
    aux = k - dist(a, b);
    if (aux >= 0 && aux % 2 == 0) {
      cout << "YES\n";
      continue;
    }
    aux = k - 1 - dist(a, x) - dist(y, b);
    if (aux >= 0 && aux % 2 == 0) {
      cout << "YES\n";
      continue;
    }
    aux -= cyclen;
    if (aux >= 0 && aux % 2 == 0) {
      cout << "YES\n";
      continue;
    }
    aux = k - 1 - dist(a, y) - dist(x, b);
    if (aux >= 0 && aux % 2 == 0) {
      cout << "YES\n";
      continue;
    }
    aux -= cyclen;
    if (aux >= 0 && aux % 2 == 0) {
      cout << "YES\n";
      continue;
    }
    cout << "NO\n";
  }
}
