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;
inline long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
void put1() { puts("YES"); }
void put2() { puts("NO"); }
void put3() { puts("-1"); }
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
using namespace std;
const int manx = 3e5 + 5;
long long head[manx], d[manx], f[manx][21], lg[manx];
long long k = 0;
struct node {
  long long v, next;
} e[manx * 2];
void add(long long u, long long v) {
  e[++k].v = v;
  e[k].next = head[u];
  head[u] = k;
}
void dfs(long long u, long long fa) {
  d[u] = d[fa] + 1;
  f[u][0] = fa;
  for (int i = 1; i <= lg[d[u]]; i++) f[u][i] = f[f[u][i - 1]][i - 1];
  for (int i = head[u]; i; i = e[i].next)
    if (e[i].v != fa) dfs(e[i].v, u);
}
long long lca(long long a, long long b) {
  if (d[a] < d[b]) swap(a, b);
  while (d[a] > d[b]) a = f[a][lg[d[a] - d[b]] - 1];
  if (a == b) return a;
  for (int i = lg[d[a]] - 1; i >= 0; i--)
    if (f[a][i] != f[b][i]) a = f[a][i], b = f[b][i];
  return f[a][0];
}
long long dis(long long x, long long y) {
  return d[x] + d[y] - d[lca(x, y)] * 2;
}
bool check(long long x, long long y) { return y >= x && x % 2 == y % 2; }
int main() {
  long long n = read();
  for (int i = 1; i < n; i++) {
    long long u = read(), v = read();
    add(u, v);
    add(v, u);
  }
  for (int i = 1; i <= n; i++) lg[i] = lg[i - 1] + (1 << lg[i - 1] == i);
  dfs(1, 0);
  long long m = read();
  for (int i = 1; i <= m; i++) {
    long long x = read(), y = read(), a = read(), b = read(), k = read();
    long long d1 = dis(a, b), d2 = dis(a, x) + 1 + dis(y, b),
              d3 = dis(a, y) + 1 + dis(x, b);
    if (check(d1, k) || check(d2, k) || check(d3, k))
      put1();
    else
      put2();
  }
  return 0;
}
