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 = 1e5 + 10;
const int logN = 17;
int n, q, h[N], par[logN][N];
vector<int> adj[N];
void Dfs(int u, int p) {
  for (int i = 1; i <= logN - 1; i++) par[i][u] = par[i - 1][par[i - 1][u]];
  for (int v : adj[u])
    if (v != p) {
      h[v] = h[u] + 1;
      par[0][v] = u;
      Dfs(v, u);
    }
}
int LCA(int u, int v) {
  if (h[u] > h[v]) swap(u, v);
  for (int i = logN - 1; i >= 0; i--)
    if (h[u] <= h[v] - (1 << i)) v = par[i][v];
  if (v == u) return u;
  for (int i = logN - 1; i >= 0; i--)
    if (par[i][u] != par[i][v]) {
      u = par[i][u];
      v = par[i][v];
    }
  return par[0][u];
}
int Dist(int u, int v) {
  int lca = LCA(u, v);
  return h[u] + h[v] - 2 * h[lca];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  Dfs(1, 1);
  scanf("%d", &q);
  while (q--) {
    int x, y, a, b, k;
    scanf("%d %d %d %d %d", &x, &y, &a, &b, &k);
    int tmp = Dist(a, b);
    if (k >= tmp && (k - tmp) % 2 == 0) {
      printf("YES\n");
      continue;
    }
    tmp = Dist(a, x) + 1 + Dist(y, b);
    if (k >= tmp && (k - tmp) % 2 == 0) {
      printf("YES\n");
      continue;
    }
    tmp = Dist(a, y) + 1 + Dist(x, b);
    if (k >= tmp && (k - tmp) % 2 == 0) {
      printf("YES\n");
      continue;
    }
    printf("NO\n");
  }
}
