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 = 1e5 + 10;
const int mo = 1e9 + 7;
int qpow(int a, int b) {
  int ans = 1;
  for (; b >= 1; b >>= 1, a = (long long)a * a % mo)
    if (b & 1) ans = (long long)ans * a % mo;
  return ans;
}
int q, n;
int f[N][21], d[N], dl[N], dr[N], idx;
vector<int> G[N];
void dfs(int u, int _f) {
  d[u] = d[f[u][0] = _f] + 1;
  dl[u] = ++idx;
  for (int i = 1; i <= 20; i++) f[u][i] = f[f[u][i - 1]][i - 1];
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v != _f) dfs(v, u);
  }
}
int lca(int u, int v) {
  if (d[u] < d[v]) swap(u, v);
  int c = d[u] - d[v];
  for (int i = 20; ~i; i--)
    if (c >> i & 1) u = f[u][i];
  if (u == v) return u;
  for (int i = 20; ~i; i--)
    if (f[u][i] ^ f[v][i]) u = f[u][i], v = f[v][i];
  return f[u][0];
}
int dis(int u, int v) { return d[u] + d[v] - 2 * d[lca(u, v)]; }
bool ok(int x, int y) { return x <= y && (y - x) % 2 == 0; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 0);
  scanf("%d", &q);
  for (int i = 1; i <= q; i++) {
    int x, y, a, b, k;
    scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
    int L = dis(x, y), d = dis(a, b);
    bool tag = 0;
    if (ok(d, k)) tag = 1;
    if (ok(dis(a, x) + dis(b, y) + 1, k)) tag = 1;
    if (ok(dis(a, y) + dis(b, x) + 1, k)) tag = 1;
    puts(tag ? "YES" : "NO");
  }
  return 0;
}
