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;
const int inf = 1e9;
int f[18][N], h[N];
vector<int> p[N];
void dfs(int x) {
  for (auto& i : p[x])
    if (i != f[0][x]) {
      f[0][i] = x;
      h[i] = h[x] + 1;
      dfs(i);
    }
}
int lca(int x, int y) {
  if (h[x] < h[y]) swap(x, y);
  for (auto i = (17); i >= (0); --i)
    if (((h[x] - h[y]) >> i) & 1) x = f[i][x];
  if (x == y) return x;
  for (auto i = (17); i >= (0); --i)
    if (f[i][x] != f[i][y]) x = f[i][x], y = f[i][y];
  return f[0][x];
}
int dis(int x, int y) { return h[x] + h[y] - h[lca(x, y)] * 2; }
bool ok(int x, int y) { return y >= x && x % 2 == y % 2; }
int main() {
  int n, q;
  scanf("%d", &n);
  for (auto i = (2); i <= (n); ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    p[x].push_back(y);
    p[y].push_back(x);
  }
  dfs(1);
  for (auto i = (1); i <= (17); ++i)
    for (auto j = (1); j <= (n); ++j) f[i][j] = f[i - 1][f[i - 1][j]];
  scanf("%d", &q);
  for (auto i = (1); i <= (q); ++i) {
    int x, y, a, b, k;
    scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
    printf("%s\n", (ok(dis(a, b), k) || ok(dis(a, x) + dis(b, y) + 1, k) ||
                    ok(dis(a, y) + dis(b, x) + 1, k))
                       ? "YES"
                       : "NO");
  }
}
