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;
vector<int> G[200005];
int n;
int f[200005], size[200005], son[200005], top[200005], d[200005];
void dfs1(int u, int pre, int deep) {
  f[u] = pre;
  d[u] = deep;
  size[u] = 1;
  for (auto v : G[u]) {
    if (v == pre) continue;
    dfs1(v, u, deep + 1);
    size[u] += size[v];
    if (size[son[u]] < size[v]) son[u] = v;
  }
}
void dfs2(int u, int tp) {
  top[u] = tp;
  if (son[u]) dfs2(son[u], tp);
  for (auto v : G[u]) {
    if (v == son[u] || v == f[u]) continue;
    dfs2(v, v);
  }
}
int LCA(int u, int v) {
  while (top[u] != top[v]) {
    if (d[top[u]] < d[top[v]]) swap(u, v);
    u = f[top[u]];
  }
  if (d[u] > d[v]) swap(u, v);
  return u;
}
int main() {
  cin >> 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);
  }
  dfs1(1, 0, 0);
  dfs2(1, 1);
  int q;
  cin >> q;
  while (q--) {
    int x, y, a, b, k;
    cin >> x >> y >> a >> b >> k;
    int lca1 = LCA(a, b);
    int dis1 = d[a] + d[b] - 2 * d[lca1];
    if (k >= dis1 && (k - dis1) % 2 == 0) {
      puts("YES");
      continue;
    }
    int lca2 = LCA(a, x), lca3 = LCA(b, y);
    int lca4 = LCA(a, y), lca5 = LCA(b, x);
    int dis2 = d[a] + d[x] - 2 * d[lca2];
    dis2 += d[b] + d[y] - 2 * d[lca3] + 1;
    int dis3 = d[a] + d[y] - 2 * d[lca4];
    dis3 += d[b] + d[x] - 2 * d[lca5] + 1;
    if (k >= dis2 && (k - dis2) % 2 == 0) {
      puts("YES");
      continue;
    }
    if (k >= dis3 && (k - dis3) % 2 == 0) {
      puts("YES");
      continue;
    }
    puts("NO");
  }
}
