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 = 3e5 + 5;
vector<int> G[N];
int pos[N], dep[N], rmq[22][2 * N], cur, n;
void dfs(int u, int fa = 0) {
  dep[u] = dep[fa] + 1;
  pos[u] = ++cur;
  rmq[0][cur] = u;
  for (int v : G[u])
    if (v != fa) {
      dfs(v, u);
      rmq[0][++cur] = u;
    }
}
int Min(int x, int y) { return dep[x] < dep[y] ? x : y; }
void rmq_init() {
  for (int l = 1, j = 1; l <= cur; l <<= 1, j++)
    for (int i = 1; i + (l << 1) - 1 <= cur; i++)
      rmq[j][i] = Min(rmq[j - 1][i], rmq[j - 1][i + l]);
}
void init() {
  dfs(1);
  rmq_init();
}
int lca(int x, int y) {
  x = pos[x], y = pos[y];
  if (x > y) swap(x, y);
  int logl = log2(y - x + 1);
  int l = 1 << logl;
  return Min(rmq[logl][x], rmq[logl][y - l + 1]);
}
int query(int x, int y) { return dep[x] + dep[y] - 2 * dep[lca(x, y)]; };
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  init();
  int q;
  scanf("%d", &q);
  while (q--) {
    int c, d, a, b, k;
    scanf("%d%d%d%d%d", &c, &d, &a, &b, &k);
    int t = query(a, b);
    if (t <= k && (k % 2 == t % 2)) {
      printf("YES\n");
      continue;
    }
    int s = query(a, c) + 1 + query(d, b);
    if (s <= k && (s % 2 == k % 2)) {
      printf("YES\n");
      continue;
    }
    int r = query(a, d) + 1 + query(c, b);
    if (r <= k && (r % 2 == k % 2)) {
      printf("YES\n");
      continue;
    }
    printf("NO\n");
  }
  return 0;
}
