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;
long long t, n, m, k, a, b, c, tmp, ans, xx, yy, gg[200005], root[200005],
    rt[200005], anx[200005], any[200005], dep[600000], jp[600000][50];
vector<int> ee[200005];
priority_queue<int> q, qq;
void jpp(int first, int fa) {
  int ii;
  gg[first] = 1;
  dep[first] = dep[fa] + 1;
  for (int ii = 1; ii <= 20; ii++) {
    jp[first][ii] = jp[jp[first][ii - 1]][ii - 1];
  }
  for (int ii = 0; ii < ee[first].size(); ii++)
    if (ee[first][ii] != fa) {
      jp[ee[first][ii]][0] = first;
      jpp(ee[first][ii], first);
    }
}
int lca(int first, int second) {
  int cost = 0, ii;
  if (dep[first] < dep[second]) swap(first, second);
  for (int ii = 20; ii >= 0; ii--) {
    if (dep[jp[first][ii]] >= dep[second]) {
      cost += (1 << ii);
      first = jp[first][ii];
    }
    if (first == second) return cost;
  }
  for (int ii = 20; ii >= 0; ii--) {
    if (jp[first][ii] != jp[second][ii]) {
      cost += (1 << ii) * 2;
      first = jp[first][ii];
      second = jp[second][ii];
    }
  }
  return cost + 2;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &a, &b);
    ee[a].push_back(b);
    ee[b].push_back(a);
  }
  for (int i = 1; i <= n; i++)
    if (ee[i].size() == 1 && gg[i] == 0) {
      jpp(i, i);
    }
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d%d%d", &xx, &yy, &a, &b, &c);
    ans = 0;
    tmp = lca(a, b);
    if (tmp <= c && (c - tmp) % 2 == 0) ans = 1;
    tmp = lca(a, xx) + lca(yy, b) + 1;
    if (tmp <= c && (c - tmp) % 2 == 0) ans = 1;
    tmp = lca(a, yy) + lca(xx, b) + 1;
    if (tmp <= c && (c - tmp) % 2 == 0) ans = 1;
    if (ans)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}
