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 P = 1e9 + 7, INF = 0x3f3f3f3f;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long qpow(long long a, long long n) {
  long long r = 1 % P;
  for (a %= P; n; a = a * a % P, n >>= 1)
    if (n & 1) r = r * a % P;
  return r;
}
long long inv(long long first) {
  return first <= 1 ? 1 : inv(P % first) * (P - P / first) % P;
}
inline int rd() {
  int first = 0;
  char p = getchar();
  while (p < '0' || p > '9') p = getchar();
  while (p >= '0' && p <= '9') first = first * 10 + p - '0', p = getchar();
  return first;
}
const int N = 1e6 + 50;
int n, m, s, sz[N], dep[N];
int L[N], R[N], fa[N], son[N], top[N];
vector<int> g[N];
void dfs(int first, int d, int f) {
  sz[first] = 1, fa[first] = f, dep[first] = d;
  for (int second : g[first])
    if (second != f) {
      dfs(second, d + 1, first), sz[first] += sz[second];
      if (sz[second] > sz[son[first]]) son[first] = second;
    }
}
void dfs(int first, int tf) {
  top[first] = tf;
  if (son[first]) dfs(son[first], tf);
  for (int second : g[first])
    if (!top[second]) dfs(second, second);
}
int lca(int first, int second) {
  while (top[first] != top[second]) {
    if (dep[top[first]] < dep[top[second]]) swap(first, second);
    first = fa[top[first]];
  }
  return dep[first] < dep[second] ? first : second;
}
int dis(int first, int second) {
  return dep[first] + dep[second] - 2 * dep[lca(first, second)];
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0, 0), dfs(1, 1);
  int q;
  scanf("%d", &q);
  while (q--) {
    int first, second, a, b, k;
    scanf("%d%d%d%d%d", &first, &second, &a, &b, &k);
    int d1 = dis(a, b);
    if (d1 <= k && d1 % 2 == k % 2) {
      puts("YES");
      continue;
    }
    int d2 = dis(a, first) + 1 + dis(second, b);
    if (d2 <= k && d2 % 2 == k % 2) {
      puts("YES");
      continue;
    }
    int d3 = dis(a, second) + 1 + dis(first, b);
    if (d3 <= k && d3 % 2 == k % 2) {
      puts("YES");
      continue;
    }
    if (d2 > d3) swap(d2, d3);
    int c = d3 - d1;
    if (c % 2 == 1) {
      puts("NO");
      continue;
    }
    if (k - d1 >= c && (k - d1 - c) % 2 == 0 ||
        k - d2 >= c && (k - d2 - c) % 2 == 0)
      puts("YES");
    else
      puts("NO");
  }
}
