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 = 200000 + 5;
int n, d[N], f[N][25], v[N], lg, k, m, tot, head[N], a, b;
void read(int &x) {
  long long f = 0;
  x = 0;
  char ch = getchar();
  for (; !(ch >= '0' && ch <= '9'); ch = getchar()) f |= ch == '-';
  for (; (ch >= '0' && ch <= '9'); ch = getchar())
    x = (x << 3) + (x << 1) + (ch ^ 48);
  x = (f) ? -x : x;
}
struct data {
  int to, next;
} e[N];
void add(int x, int y) {
  e[++tot].to = y;
  e[tot].next = head[x];
  head[x] = tot;
}
void dfs(int x) {
  v[x] = 1;
  for (int i = head[x]; i; i = e[i].next) {
    int y = e[i].to;
    if (v[y]) continue;
    d[y] = d[x] + 1;
    f[y][0] = x;
    dfs(y);
  }
}
int lca(int x, int y) {
  if (d[x] < d[y]) swap(x, y);
  for (int i = lg; i >= 0; i--)
    if (d[f[x][i]] >= d[y]) x = f[x][i];
  if (x == y) return y;
  for (int i = lg; i >= 0; i--)
    if (f[x][i] != f[y][i]) x = f[x][i], y = f[y][i];
  return f[x][0];
}
bool q(int g) {
  if (g <= k) return (g % 2 == k % 2) ? 1 : 0;
  return 0;
}
int ro(int x, int y) {
  int root = lca(x, y);
  return (d[x] - d[root]) + (d[y] - d[root]);
}
int main() {
  read(n);
  for (int i = 1; i < n; i++) {
    int x, y;
    read(x), read(y);
    add(x, y);
    add(y, x);
  }
  d[1]++;
  f[1][0] = 1;
  dfs(1);
  lg = 20;
  for (int j = 1; j <= lg; j++)
    for (int i = 1; i <= n; i++) f[i][j] = f[f[i][j - 1]][j - 1];
  read(m);
  for (int i = 1; i <= m; i++) {
    int x, y;
    read(x), read(y), read(a), read(b), read(k);
    int fi = ro(a, b);
    int s = ro(a, x) + ro(y, b) + 1;
    int t = ro(a, y) + ro(x, b) + 1;
    if (q(fi) || q(s) || q(t))
      printf("YES\n");
    else
      printf("NO\n");
  }
}
