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 maxn = 100000;
int n, N, q;
int f[maxn + 5][25];
int dis[maxn + 5];
vector<int> E[maxn + 5];
int get_log(int x) {
  int t = 0;
  while ((1 << (t + 1)) <= x) t++;
  return t;
}
void dfs(int x, int fa) {
  f[x][0] = fa;
  dis[x] = dis[fa] + 1;
  for (int i = 0; i < E[x].size(); i++) {
    int v = E[x][i];
    if (v != fa) {
      dfs(v, x);
    }
  }
}
int get_lca(int u, int v) {
  if (dis[u] < dis[v]) swap(u, v);
  int d = dis[u] - dis[v];
  for (int i = 0; i <= N; i++) {
    if ((1 << i) & d) {
      u = f[u][i];
    }
  }
  if (u == v) return u;
  for (int i = N; i >= 0; i--) {
    if (f[u][i] != f[v][i]) {
      u = f[u][i];
      v = f[v][i];
    }
  }
  return f[u][0];
}
int main() {
  scanf("%d", &n);
  N = get_log(n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    E[u].push_back(v);
    E[v].push_back(u);
  }
  dis[0] = 0;
  dfs(1, 0);
  for (int j = 1; j <= N; j++)
    for (int i = 1; i <= n; i++) f[i][j] = f[f[i][j - 1]][j - 1];
  scanf("%d", &q);
  while (q--) {
    bool flag = false;
    int x, y, a, b, k, u, v;
    scanf("%d%d%d%d%d", &x, &y, &a, &b, &k);
    u = get_lca(a, b);
    int len = dis[a] + dis[b] - 2 * dis[u];
    if ((len & 1) == (k & 1) && (len <= k)) flag = true;
    if (!flag) {
      u = get_lca(a, x);
      v = get_lca(b, y);
      int len1 =
          dis[a] + dis[x] - 2 * dis[u] + dis[b] + dis[y] - 2 * dis[v] + 1;
      u = get_lca(a, y);
      v = get_lca(b, x);
      int len2 =
          dis[a] + dis[y] - 2 * dis[u] + dis[b] + dis[x] - 2 * dis[v] + 1;
      if (((len1 & 1) == (k & 1) && len1 <= k) ||
          ((len2 & 1) == (k & 1) && len2 <= k))
        flag = true;
    }
    if (flag)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}
