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>
int n, q;
int d[100005], lg[100005], f[100005][25];
std::vector<int> v[100005];
void dfs(int x, int fa) {
  f[x][0] = fa;
  d[x] = d[fa] + 1;
  for (int i = 1; i <= lg[d[x]]; ++i) f[x][i] = f[f[x][i - 1]][i - 1];
  for (int i : v[x])
    if (i != fa) dfs(i, x);
}
inline int lca(int x, int y) {
  if (d[x] < d[y]) std::swap(x, y);
  while (d[x] > d[y]) x = f[x][lg[d[x] - d[y]]];
  if (x == y) return x;
  for (int i = lg[d[x]]; ~i; --i)
    if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
  return f[x][0];
}
inline int dis(int x, int y) { return d[x] + d[y] - (d[lca(x, y)] << 1); }
int main() {
  scanf("%d", &n);
  for (int i = 1, x, y; i < n; ++i) {
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (int i = 2; i <= n; ++i) lg[i] = lg[i >> 1] + 1;
  dfs(1, 0);
  scanf("%d", &q);
  for (int i = 0, x, y, a, b, k, Dis; i < q; ++i) {
    scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
    Dis = dis(a, b);
    if (Dis <= k && (Dis & 1) == (k & 1)) {
      puts("YES");
      continue;
    }
    Dis = dis(a, x) + dis(b, y) + 1;
    if (Dis <= k && (Dis & 1) == (k & 1)) {
      puts("YES");
      continue;
    }
    Dis = dis(a, y) + dis(b, x) + 1;
    if (Dis <= k && (Dis & 1) == (k & 1)) {
      puts("YES");
      continue;
    }
    puts("NO");
  }
  return 0;
}
