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;
int qur;
bool vis[200001];
int f[200001][21], deep[200001];
int first[200001], nex[200001], v[200001], num;
inline void add(int x, int y) {
  num++;
  v[num] = y;
  nex[num] = first[x];
  first[x] = num;
}
inline void pre(int u, int fa) {
  vis[u] = true;
  deep[u] = deep[fa] + 1;
  f[u][0] = fa;
  for (int i = 1; i <= 20; i++) {
    f[u][i] = f[f[u][i - 1]][i - 1];
  }
  for (int i = first[u]; i != -1; i = nex[i]) {
    int to = v[i];
    if (!vis[to]) pre(to, u);
  }
}
inline int lca(int x, int y) {
  if (deep[x] < deep[y]) swap(x, y);
  for (int i = 20; i >= 0; i--) {
    if (deep[f[x][i]] >= deep[y]) {
      x = f[x][i];
    }
  }
  if (x == y) return x;
  for (int i = 20; i >= 0; i--) {
    if (f[x][i] != f[y][i]) {
      x = f[x][i];
      y = f[y][i];
    }
  }
  return f[x][0];
}
int main() {
  memset(first, -1, sizeof first);
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    add(x, y);
    add(y, x);
  }
  pre(1, 0);
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) {
    int x, y, a, b, k;
    bool flag = false;
    scanf("%d %d %d %d %d", &x, &y, &a, &b, &k);
    int len1 = deep[a] + deep[b] - 2 * deep[lca(a, b)];
    int len2 = deep[a] + deep[x] - 2 * deep[lca(a, x)] + deep[y] + deep[b] -
               2 * deep[lca(y, b)] + 1;
    int len3 = deep[a] + deep[y] - 2 * deep[lca(a, y)] + 1 + deep[x] + deep[b] -
               2 * deep[lca(x, b)];
    if (len1 <= k && (k - len1) % 2 == 0) {
      flag = true;
    }
    if (len2 <= k && (k - len2) % 2 == 0) {
      flag = true;
    }
    if (len3 <= k && (k - len3) % 2 == 0) {
      flag = true;
    }
    if (flag)
      puts("YES");
    else
      puts("NO");
  }
  return 0;
}
