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;
int n, m, f[100100][30], dep[100100];
int cnt, bg[100100], ne[200100], to[200100];
void add(int x, int y) {
  cnt++;
  ne[cnt] = bg[x];
  to[cnt] = y;
  bg[x] = cnt;
}
void dfs(int x, int fa) {
  f[x][0] = fa;
  dep[x] = dep[fa] + 1;
  for (int i = bg[x]; i; i = ne[i]) {
    int y = to[i];
    if (y != fa) {
      dfs(y, x);
    }
  }
}
void ycl() {
  for (int j = 1; j <= 25; j++) {
    for (int i = 1; i <= n; i++) {
      f[i][j] = f[f[i][j - 1]][j - 1];
    }
  }
}
int get_le(int x, int y) {
  if (dep[x] < dep[y]) {
    swap(x, y);
  }
  int sum = 0;
  for (int i = 25; i >= 0; i--) {
    if (dep[f[x][i]] >= dep[y]) {
      sum += (1 << i);
      x = f[x][i];
    }
  }
  if (x == y) {
    return sum;
  }
  for (int i = 25; i >= 0; i--) {
    if (f[x][i] != f[y][i]) {
      x = f[x][i], y = f[y][i];
      sum += (1 << i) * 2;
    }
  }
  return sum + 2;
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  dfs(1, 0);
  ycl();
  cin >> m;
  for (int i = 1; i <= m; i++) {
    int x, y, a, b, k;
    scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
    int q1, q2, q3;
    q1 = get_le(a, b);
    q2 = get_le(a, x) + get_le(b, y) + 1;
    q3 = get_le(b, x) + get_le(a, y) + 1;
    if ((q1 <= k && q1 % 2 == k % 2) || (q2 <= k && q2 % 2 == k % 2) ||
        (q3 <= k && q3 % 2 == k % 2)) {
      puts("YES");
    } else {
      puts("NO");
    }
  }
  return 0;
}
