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 = 200100, M = 20010010;
int head[N], ver[M], Next[M], tot;
int n, q;
int f[N][20], d[N], t, dist[N];
void add(int x, int y) {
  ver[++tot] = y;
  Next[tot] = head[x];
  head[x] = tot;
}
void bfs() {
  queue<int> q;
  q.push(1);
  d[1] = 1;
  while (q.size()) {
    int x = q.front();
    q.pop();
    for (int i = head[x]; i; i = Next[i]) {
      int y = ver[i];
      if (d[y]) continue;
      d[y] = d[x] + 1;
      f[y][0] = x;
      for (int j = 1; j <= t; j++) f[y][j] = f[f[y][j - 1]][j - 1];
      q.push(y);
    }
  }
}
int lca(int x, int y) {
  if (d[x] > d[y]) swap(x, y);
  for (int i = t; i >= 0; i--)
    if (d[f[y][i]] >= d[x]) y = f[y][i];
  if (x == y) return x;
  for (int i = t; i >= 0; i--)
    if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
  return f[x][0];
}
int dis(int u, int v) {
  int dx = lca(u, v);
  return d[u] + d[v] - 2 * d[dx];
}
int main() {
  scanf("%d", &n);
  tot = 0;
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  t = (int)(log(n) / log(2) + 1);
  bfs();
  scanf("%d", &q);
  while (q--) {
    int x, y, a, b, k;
    scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
    int s1 = dis(a, b), s2 = dis(a, x) + dis(b, y) + 1,
        s3 = dis(a, y) + dis(b, x) + 1;
    if ((k >= s1 && (k - s1) % 2 == 0) || (k >= s2 && (k - s2) % 2 == 0) ||
        (k >= s3 && (k - s3) % 2 == 0))
      printf("YES\n");
    else
      printf("NO\n");
  }
}
