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 INF = 0x3f3f3f3f;
const int N = 1e5;
vector<int> v[N + 5];
int vis[N + 5];
int d[N + 5], f[N + 5][50];
int n;
int t;
queue<int> q;
void bfs() {
  q.push(1);
  d[1] = 1;
  while (q.size()) {
    int x = q.front();
    q.pop();
    for (auto y : v[x]) {
      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 main() {
  scanf("%d", &n);
  t = (int)(log(n) / log(2.0)) + 1;
  for (int i = 1; i < n; i++) {
    int uu, vv;
    scanf("%d%d", &uu, &vv);
    v[uu].push_back(vv);
    v[vv].push_back(uu);
  }
  bfs();
  int q;
  scanf("%d", &q);
  while (q--) {
    int x, y, a, b, k;
    scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
    int dis1 = d[a] + d[b] - 2 * d[lca(a, b)];
    int dis2 =
        d[a] + d[x] - 2 * d[lca(x, a)] + d[b] + d[y] - 2 * d[lca(y, b)] + 1;
    int dis3 =
        d[a] + d[y] - 2 * d[lca(y, a)] + d[b] + d[x] - 2 * d[lca(x, b)] + 1;
    bool f = false;
    if (k >= dis1) {
      if ((k - dis1) % 2 == 0) f = true;
    }
    if (k >= dis2) {
      if ((k - dis2) % 2 == 0) f = true;
    }
    if (k >= dis3) {
      if ((k - dis3) % 2 == 0) f = true;
    }
    if (f)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}
