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 MAX = 1e5 + 10;
const int MAX2 = 17;
vector<vector<int>> g(MAX);
int n, p;
int pai[MAX2][MAX];
int in[MAX], out[MAX];
int sz[MAX];
void dfs(int k, int pr) {
  in[k] = p++;
  sz[k] = pr;
  for (int i = 0; i < (int)g[k].size(); i++) {
    if (in[g[k][i]] == -1) {
      pai[0][g[k][i]] = k;
      dfs(g[k][i], pr + 1);
    }
  }
  out[k] = p++;
}
void build(int raiz) {
  for (int i = 0; i < n; i++) pai[0][i] = i;
  p = 0, memset(in, -1, sizeof(in));
  dfs(raiz, 0);
  for (int k = 1; k < MAX2; k++)
    for (int i = 0; i < n; i++) pai[k][i] = pai[k - 1][pai[k - 1][i]];
}
bool anc(int a, int b) { return in[a] <= in[b] and out[a] >= out[b]; }
int lca(int a, int b) {
  if (anc(a, b)) return a;
  if (anc(b, a)) return b;
  for (int k = MAX2 - 1; k >= 0; k--) {
    if (!anc(pai[k][a], b)) a = pai[k][a];
  }
  return pai[0][a];
}
int dist(int a, int b) { return sz[a] + sz[b] - 2 * sz[lca(a, b)]; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  build(0);
  int m;
  cin >> m;
  for (int i = 0; i < m; i++) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    x--, y--, a--, b--;
    int d1 = dist(a, b);
    int d2 = dist(a, x) + 1 + dist(y, b);
    int d3 = dist(a, y) + 1 + dist(x, b);
    if (d1 <= k and (k - d1) % 2 == 0)
      cout << "YES" << '\n';
    else if (d2 <= k and (k - d2) % 2 == 0)
      cout << "YES" << '\n';
    else if (d3 <= k and (k - d3) % 2 == 0)
      cout << "YES" << '\n';
    else
      cout << "NO" << '\n';
  }
  exit(0);
}
